# Finding signed areas and volumes inspired by technology.

Abstract

We describe how we can find the singed area bounded by a parametric curse with respect to a slanted line, y = mx + b, in two dimensions. We extend this idea to the corresponding setting in three dimensions. These are special cases of Green's Theorem and the Divergence Theorem, but the ideas and methods introduced here are accessible to a wider audience.

1 Introduction

In a calculus textbook (see  or ), the Riemann integral [[integral].sup.b.sub.a] f(x)dx for a continuous function f over the interval [a, b] represents the net area of the region bounded by the curve y = f(x), two vertical lines x = a and x = b (both perpendicular to the x-axis). and the x-axis. In this paper, we discuss ways of finding the area bounded by a parametric curve [x(t), y(t)], a slanted line y = mx + b, and the perpendiculars to the line from (x([t.sub.1]), y([t.sub.1])) to (x([t.sub.2]); y([t.sub.2])) in two dimensions. Since we replace the graph of a function y = f(x) with a parametric curve, the sign of the bounded area depends on the direction of the given parametric curve. Throughout this paper, we will use the terminology of finding the net signed area (or simply area) bounded by a parametric curve [x(t); y(t)] with respect to a slanted line y = mx + b without causing any confusion. We generalize the idea from two dimensions (2-D) to three dimensions (3-D) and use similar terminology throughout the paper.

In Section l, we illustrate how technology has inspired us to conceive formulae in two dimensions. In Section 2, we generate the 2-D formulae to corresponding ones in 3-D. In Section 3, we describe a special method of finding the volume bounded by a surface satisfying the form F(x, y, z) and a general plane. We do not require readers to have knowledge of Green's Theorem or the Divergence Theorem. However, we remind those readers who are familiar with Green's Theorem and the Divergence Theorem that the results derived in this paper are consistent with those of Green's Theorem and the Divergence Theorem.

1.1 Motivation

We intend to find the area of one of the regions bounded by the curve C : [x.sup.2] + [y.sup.2] + 2.22x - 0.1y - 0.529 = 0 and the line [??] : y = 0.9802x + 1.866, which is shown in Figure 1. We note that the curve C and the line [??] were drawn arbitrarily employing Dynamic Geometry software [C1assPad Manager]. The line [??} divides the circle into two portions, we select the smaller portion (see Figure 1) for demonstration. We select such C because if we were to apply traditional techniques using either vertical or horizontal partitions (dx or dy respectively), we would need to divide the area into two sub-areas since a vertical or horizontal line intersects C at two different intersections. However, if we think of EF as our new x-axis and choose one line that is perpendicular to EF as our new y-axis, then we avoid multiple partitions and can apply a numerical approximation technique to approximate the Riemann integral for the enclosed region before analytically finding the bounded area.

[FIGURE 1 OMITTED]

We will use ClassPad (see ) to describe how we apply the Trapezoidal Rule to approximate the area of the region.

* We note the distance of EF equals 2.444009. Subdivide EF into 19 panels: 2.444009/19 = 0.1286320526.

* Each subinterval is of length 0.1286320526.

* The length of HC is tabled below:

[TABLE 1 OMITTED]

We calculate the area of each trapezoidal region and table them below. Note that column represents the height HG and column B represents the area of each trapezoidal region. For example the cell B1 = (A1 + A2) x (0.1286320526)/2 is the area of the first trapezoid.

[TABLE 2 OMITTED]

* We sum up the values in column B and save it in cell C1 = 1.4191, which is the approximation of the bounded area.

* The video clip and eActivity using ClassPad (see ) describing this example can be found in  and  respectively.

1.2 Theoretical Approach

A curve w (t) is smooth if w' (t) is continuous and never equal to the zero vector. We first integrate a smooth curve C of the form w(t) = [x(t), y(t)], where [t.sub.1] [less than or equal to] t [less than or equal to] [t.sub.2], with respect to a line L : y = mx through the origin. We remark that the curve C will travel from t = [t.sub.1] to t = [t.sub.2]; this becomes important when we interpret the area, which will have positive or negative value. For example, we sketch the of the curve C (that is traveling from P to Q), and the line segment AB, which is y = mx, in Figure 2 below. We note that PA and QB are both perpendicular to the line segment AB.

[FIGURE 2 OMITTED]

Let [B.sub.1] = {[e.sub.1], [e.sub.2]} be the standard basis for [[??].sup.2,] and [B.sub.2] = {[c.sub.1], [c.sub.2]} be another basis for [[??].sup.2], where [c.sub.1] = (cos [theta], sin [theta]) is a unit direction vector of the line L and [c.sub.2] = (-sin [theta], cos [theta]) is a unit vector perpendicular to [c.sub.1] with [theta] = [tan.sup.-1] m.

We first express [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] as a vector relative to the basis [B.sub.2]. In other words, we need to discover [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] so that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], or

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1)

This implies that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (2)

The integral of [x(t), y(t)] with respect to a line L : y = mx through the origin is

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (3)

Next, integrate a parametric curve w(t) = [x(t), y(t)], [t.sub.1] [less than or equal to] t [less than or equal to] [t.sub.2], with respect to a line L : y = mx + b (let [theta] = [tan.sup.-1] m). By shifting the bounded region vertically by b. we can then apply the same concept above to find the coordinate vector [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] of [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] relative to [B.sub.2] below.

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (4)

Therefore, the integral of w(t) = [x(t), y(t)], [t.sub.1] [less than or equal to] t [less than or equal to] [t.sub.2], with respect to the line L : y = mx + b (let [theta] = [tan.sup.-1] m) is

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (5)

To restate what we have just proved:

Theorem 1 Let C he the smooth curve w(t) = [x(t), y(t)], where [t.sub.1] [less than or equal to] t [less than or equal to] [t.sub.2]. Let R he the region hounded by C, the line y = mx + b, and the perpendiculars to the line from (x([t.sub.1]), y([t.sub.1])) to (x([t.sub.2]), y([t.sub.2])). Then the area of R is given by,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (6)

Green's Theorem in our discussion case be summarized as follows:

Theorem 2 Let C be the smooth curve w(t) = [x(t), y(t)], where [t.sub.1] [less than or equal to] t [less than or equal to] [t.sub.2]. Let R be the region bounded by C, the line y = mx + b, and the perpendiculars to the line from (x([t.sub.1]), y([t.sub.1])) to (x([t.sub.2]), y([t.sub.2])). We denote the counterclockwise boundary curve of R by [partial derivative]R. If P and Q are scalar fields with continuous partial derivatives satisfying [partial derivative]Q/[partial derivative]x - [partial derivative]P/[partial derivative]y = 1, then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (7)

Note. For convenience, we choose P(x, y) = -y/2 and Q(x, y) = x/2.

Remark 3 Observe Theorem 1 proffers an explicit formula for finding the area enclosed by a parametric closed curve. Suppose the curve C is a semicircle described by the thick curve, [cos t, sin t], traveling from D to E, where t [member of] [0, [pi]]. The line L = [??] is y = x - 4, as shown in Figure 2.

[FIGURE 3 OMITTED]

The area of the the region R bounded by C, the line L, and the perpendiculars to the line L from D to E is enclosed by the closed curve D [right arrow] E [right arrow] A [right arrow] B [right arrow] D. This area is the area of the semicircle plots the area of the trapezoid DEAB. Since the closed curve is traversed counterclockwise, (6) gives us a negative value of -[pi]/2 - 4, and therefore the area is [pi]/2 + 4.

For the remainder of the paper, the Nsord "area" represents a net signed area. The area enclosed by a counterclockwise curse is negative and the area enclosed by a clockwise curse is positive due to the construction of [c.sub.1] in (1) in section 1.2.

Remark 4 Set m = b = 0 in (6), then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (8)

as expected. In addition, (7) also provides the signed area enclosed by a curve [x(t), y(t)] , a line L, and the perpendiculars to the line L from (x([t.sub.1]), y([t.sub.1])) to (x([t.sub.2]), y([t.sub.2])).

Corollary 5 Let C be a closed smooth curve traveling only in one direction, then (6) produces the area bounded by the closed curve which is independent of the location of the line L.

Remark 6 Corollary 5 and the example in Remark 3 coincide with Green's Theorem in the sense that the area enclosed by a closed piecewise smooth curve can be obtained by utilizing contour integrals along the boundary curve traversed in one direction.

1.3 Examples

Using Theorem 1, we can find the area bounded by two parametric curves without utilizing Green's Theorem.

Example 7 Given two curies, a cardioid [C.sub.1] = [2cos(t) - cos(2t), 2sin(t) - sin(2t)] and a circle [C.sub.2] = [0. 5 - 3 cos(t), 0.5 + 3 sin(t)], inhere t [member of] [0, 2[pi]].

(1) Find the area of the region that is inside the cardioid [C.sub.1] and exterior to the circle [C.sub.2]. (See Figure 3 below.)

(2) Find the area of the region that is inside both [C.sub.1] and [C.sub.2].

[FIGURE 4 OMITTED]

Part (1):

Step 1. First, we find the intersections of these two curves at

A = (-2.108615230, 1.981595956) and B = (.2016062939, -2.485421672), respectively.

Step 2. We calculate the area bounded by [C.sub.1] and line segment AB. Re integrate along [C.sub.1] counter-clockwise from A (t = 2.574088482) to B (t = 4.402664587) by applying (6) to compute the area hounded by [C.sub.1] and the line equation AB:

y = -1.931x - 2.09,

and obtain

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (9)

Note that the answer above is negative since we integrate along a counterclockwise path. Step 3. We next calculate the area bounded by [C.sub.2] and the line segment AB. We integrate along [C.sub.2] clockwise from B (t = 4.613764607) to A (t = 2.625063229) by applying (6) to compute the area bounded by [C.sub.2] and the line equation AB:

y = -1.931x - 2.09,

and obtain

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (10)

Therefore, the area bounded between [C.sub.1] and [C.sub.2] inside the cardioid [C.sub.1] and exterior to the circle [C.sub.2] is 2.701472124.

Part (2): We first find the area of the cardioid of [C.sub.1] by applying equation (8), which yields 18:84955592. Next we subtract the answer of part 1 from the area of the cardiod [C.sub.1]. Our result for part (2) is 16.14808380.

The Maple worksheet for this example can be found in .

Remark 8 Let [C.sub.1] be the path traversed along [C.sub.1] counterclockwise from A to B and [C.sub.2] be the path traversed along [C.sub.2] clockwise from B to A. Using Green's Theorem on the curve C = [C.sub.1] [[C.sub.2], we obtained 2.701472124 for the area of the region enclosed by C:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Example 9 Let the curve C be an ellipse given by [A cos t;B sin t], where t [member of] [0; 2[pi]], and a line L : y = mx + b. We obtain AB[pi] for the area of the region inside the ellipse by applying (6), regardless of whether the line L intersects the curve C.

Another example below shows that if a curve traverses in only one direction, we will obtain the area enclosed by the curve by applying (6).

Example 10 Let r(t) = sin 3t, f(t) = r(t) x cos t, g(t) = r(t) x sin t, the line L = [??] be y = -0:9x - 1:378. If C is the curve [f(t); g(t)], where t [member of] [0, [pi]/2]. Find the area bounded by C with respect to the line L.

Sketch the graph of C, line L = [??], and perpendiculars AE and BD, from A = [f(0), g(0)] and B = [f([pi]/2), g([pi]/2)] to L, respectively, in Figure 5 below. Find the areas of two portions, one being the loop going from A back to A in a counterclockwise direction and the other enclosed by ABDEA in a clockwise direction. The area of the first portion is

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

[FIGURE 5 OMITTED]

We apply formula (6) to compute the area for the second portion

1/1 + [m.sup.2] [[integral].sup.[pi]/2.sub.[pi]/3] (-x(t)m + y(t) - b)(x'(t) + y'(t)m)dt = 0.3056748917.

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Therefore, the area of the region hounded by C with respect to the line L is 0.5674742795. The corresponding Maple worksheet can be found in .

Example 11 Let r(t) = sin 3t, f(t) = r(t) x cos t, g(t) = r(t) x sin t. Furthermore, define

F(t) = f(t - [pi]/6) and

G(t) = g(t + [pi]/6).

Find the area enclosed by [F(t), G(t)], where t [member of] [0, [pi]]. Sketch the graph of [F(t), G(t)] as follows:

[FIGURE 6 OMITTED]

By Corollary 5, we can apply (8) in Remark 4 to obtain the value [pi]/8 . Note that since the curve travels only in one (clockwise) direction (we leave this verification to the reader), the value represents the area enclosed by the curve. The Maple worksheet for this example can be found in .

2 Extending ideas from 2-D to 3-D

Suppose x (s, t), y (s, t), and z (s, t) are scalar functions, and D [subset or equal to] [R.sup.2]. The set

w (s, t) = {[x (s, t) , y (s, t), z (s, t)] [member of] [R.sup.3] : (s, t) [member of] D}

is called a parametric surface or simply a surface. The parametric surface is called smooth if the tangent vectors in the s and t directions satisfy

[w.sub.s] x [w.sub.t] [not equal to] (0, 0, 0) throughout its domain. (11)

For the rest of paper, we will assume a surface is orientable; i.e., it is possible to choose a unit normal vector n at every point continuously over the surface.

We want to integrate a smooth parametric surface w(s, t) = [x(s, t), y(s, t), z(s, t)], [t.sub.1] [less than or equal to] t [less than or equal to] [t.sub.2], [s.sub.1] [less than or equal to] s [less than or equal to] [s.sub.2], with respect to a plane P passing through the origin and spanned by vectors u and v; i.e., P = {0 + au + bv : a, b [member of] R}.

Let [B.sub.1] = {[e.sub.1], [e.sub.2], [e.sub.3]} be the standard basis for [R.sup.3], and [B.sub.2] = {[p.sub.1]/[parallel][p.sub.1][parallel], [p.sub.2]/[parallel][p.sub.2][parallel], n/[parallel]n[parallel] be another ortho-normal basis for [R.sup.3]. To form [B.sub.2],

Case 1: If u [perpendicular to] v, then we choose [p.sub.1] = u, [p.sub.2], = v, and n = [p.sub.1] x [p.sub.2].

Case 2: If u is not perpendicular to v, then [p.sub.1] is the orthogonal projection of u on v, [p.sub.2] = u - [p.sub.1], and n = [p.sub.1] x [p.sub.2] is the normal vector for the plane P.

First we need to find the vector [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] so that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (12)

Therefore,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (13)

The integral of the surface w(s, t) = [x(s, t), y(s, t), z(s, t)], [t.sub.1] [less than or equal to] t [less than or equal to] [t.sub.2], [s.sub.1] [less than or equal to] s [less than or equal to] [s.sub.2], with respect to P is

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (14)

We are ready to integrate a smooth parametric surface w(s, t) = [x(s, t), y(s, t), z(s, t)], [t.sub.1] [less than or equal to] t [less than or equal to] [t.sub.2], [s.sub.1] [less than or equal to] s [less than or equal to] [s.sub.2], with respect to a general plane P = {d + au + by : a, b [member of] R} that does not pass through the origin and has only one z-intercept d. In other words, P is the plane spanned by u and v, passing through d = (0, 0, [d.sub.3]).

Using the same notation as we did previously, we adjust (13) by replacing w with w- [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (by doing this, we are shifting the surface w down vertically by [d.sub.3]) We have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (15)

Therefore, the volume bounded by the parametric surface w(s, t) = [x(s, t), y(s, t), z(s, t)], [t.sub.1] [less than or equal to] t [less than or equal to] [t.sub.2], [s.sub.1] [less than or equal to] s [less than or equal to] [s.sub.2], with respect to a general plane P = {d + au + bv : a, b [member of] R} is

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (16)

We next verify that our 3-D formula does reduce to the corresponding one in 2-D. We consider w(t) = [x(t), y(t), z(t)], [t.sub.1] [less than or equal to] t [less than or equal to] [t.sub.2].

The plane equation P = {d + au + bv : a, b [member of] R} becomes a line L = {d + au :a [member of] R} when b = 0. and the slope in of this line is [u.sub.2]/[u.sub.1], where u = ([u.sub.1], [u.sub.2]). Let [theta] = [tan.sup.- 1](m), then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

In this case, we use [p.sub.1] = u = ([u.sub.1], [u.sub.2]), and n = (-[u.sub.2], [u.sub.1]), al) (note that [p.sub.1] [perpendicular to] n and [parallel]n[parallel] = [parallel][p.sub.1][parallel] = [parallel]u[parallel]. We first find the coordinate vector [p(t), q(t)] of [x(t), y(t)] with respect to the basis {[p.sub.1]/[parallel][p.sub.1][parallel], n/[parallel]n[parallel]}, where [p.sub.1]/[parallel][p.sub.1][parallel] is an unit direction vector of the line L = {d + au :a [member of] [??]}. Let d =(0. b) be the y-intercept of the graph of L. Therefore,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (17)

which coincides with (4).

In summary, the above proof leads to:

Theorem 12 Let S be the smooth parametric surface, w(s, t) = [x(s, t), y(s, t), z(s, t)], [t.sub.1] [less than or equal to] t [less than or equal to] [t.sub.2], [s.sub.1] [less than or equal to] s [less than or equal to] [s.sub.2]. Let R be the region bounded by S, the plane P : ax + by + cz = d and the perpendicular projection from the surface to the plane P. Then the volume of R is given by

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (18)

where [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], and [p.sub.1] and [p.sub.2] are two orthonormal vectors on the plane P.

For the remainder of the paper, the word "volume" represents a signed volume, which depends on the orientation of a surface. The Divergence Theorem in our discussion can be summarized as follows:

Theorem 13 Let

F (x, y, z) = [P (x, y, z), Q (x, y, z), R (x, y, z)]

be a vector field defined on a solid region R (with div F = 1) hounded by a .smooth surface w (s, t) = {[x(s, t), y(s, t), z(s, t))] : [t.sub.1] [less than or equal to] t [less than or equal to] [t.sub.2], [s.sub.1] [less than or equal to] s [less than or equal to] [s.sub.2]}, a plane P : ax + by + cz = d, and the perpendicular projection from the surface to the plane P. Let S be the boundary surface of R and n be the outward pointing unit normal vector at each point of S. If the components of F have continuous first order partial derivatives on an open region that contains R, then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (19)

where [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], div F = [partial derivative]P/[partial derivative]x + [partial derivative]Q/[partial derivative]y + [partial derivative]R/[partial derivative]z, and [p.sub.1] and [p.sub.2] are two orthonormal vectors on the plane P.

Remark 14 Choose the plane to be the xy-plane (i.e., z = 0) and use the standard basis {[e.sub.1], [e.sub.2], [e.sub.3]} for [R.sup.3]. Then (19) reduces to

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (20)

as expected.

Corollary 15 If the smooth surface w(s, t) is an orientable closed surface, (19) produces the volume of the region bounded by the closed surface, and this volume is independent of the location of the plane P.

Remark 16 This is a simple application of the Divergence Theorem.

The next two examples demonstrate the use of Theorem 13 and Corollary 15.

Example 17 Let S be the ellipsoid [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], where s [member of] [0, 2[pi]], and t [member of] [0, [pi]], and P he the plane 2x + y + z = 0 (passing through the origin). See Figure 6 below.

[FIGURE 7 OMITTED]

Then (18) produces the volume of the ellipsoid regardless of the location of the plane P. This is a direct observation, from the Divergence Theorem. (See Maple worksheet in ).

Example 18 Let S be the ellipsoid represented by the following parametric equation

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

where s [member of] [[pi], 2[pi]]. and t [member of] [0, [pi]/2], and P be the plane x + y + z = -3. See the graph below.

[FIGURE 8 OMITTED]

Let R be the region hounded by S, the plane P, and the perpendicular projection from the surface to the plane P. Then the signed volume of R is -2[pi] + 26/9 when Theorem 12 or 13 is applied. (See Maple worksheet 11.).

3 Another Rotation and Shifting Method

It may not be trivial to find the intersection between a smooth surface and a plane. Consequently, it is difficult to find the volume of the solid bounded by the surface and the plane. In this section, we describe how we use the idea of rotation to aid us in finding the volume of the solid bounded by a closed orientable smooth surface S of the form F(x, y, z) = 0 and a plane [P.sub.1] : ax + by + cz = d. Throughout this section. we assume the plane [P.sub.1] cuts the solid bounded by the surface S into only two pieces. Our goal is to find the volumes of these two respective pieces bounded by [P.sub.1] and S. The procedure is as follows:

Step 1. We first consider the plane [P.sub.2] : ax + by + cz. = 0, which is parallel to [P.sub.1] and passes through the origin.

Step 2. We pick three points [X.sub.0], [X.sub.1], and [X.sub.2] on [P.sub.2], and form the orthonormal basis {[p.sub.1], [p.sub.2], n}, where n is the unit normal vector of [P.sub.2] (or [P.sub.1]) that points toward the the region whose volume we seek.

Step 3. We now consider the plane [P.sub.1] : ax + by + cz = d. By Theorem 12, we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], where [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], A = [[p.sub.1], [p.sub.2], n], and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. In other words, we express X in terms of [X.sup.*].

Step 4. We substitute X with [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] into the formula F(x, y, z) = 0, which results in an equation G(P, q, w) = 0.

Step 5. Employing spherical coordinates we solve the equation G(p, q, w) = G([rho]sin t cos s, [rho] sin t sin s, [rho] cos t) = 0 for [rho].

Step 6. We use the following formula to find the volume of the region bounded by [P.sub.1] and F(x, y, z) = 0:

[[integral].sup.2[pi].sup.0] [[integral].sup.[pi]/2.sup.0] ([[integral].sup.[rho](s,t).sub.0] ([[rho].sup.2] sin t) d[rho]) dtds. (21)

Note: Maple allows us to solve for [rho] above. If the exact integration value from (21) is not possible when using a CAS, we apply higher dimension Simpson's rule to obtain an approximation. We summarize the discussion above into the following:

Theorem 19 Let S be the .smooth parametric surface of the form F(x, y, z) = 0 (or we write F(X) = 0 for simplicity). Let R be the region bounded by S and the plane P : ax + by + cz = d. Furthermore, we assume the.follou1ng conditions are met:

(i) If [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], where A = [[p.sub.1], [p.sub.2], n] and [p.sub.1], [p.sub.2] are two perpendicular unit vectors on the plane P, and n is the unit normal vector of P that points toward the the region whose volume we seek.

(ii) If G([rho] sin t cos s, [rho] sin t sin s, [rho] cos t) = 0 is solvable for [rho], then the volume of R is given by

[[integral].sup.2[pi].sub.0] [[integral].sup.[pi]/2.sub.0] ([[integral].sup.[rho](s,t).sub.0] ([[rho].sup.2] sin t) d[rho]) dtds. (22)

3.1 An Example

Example 20 Given the surface S : 4[x.sup.2] + 4[y.sup.2] + 4[z.sup.2] + 4xy + 4xz + 4yz - 5 = 0 and the plane P : x + y + z - 1 = 0, find the respective volumes bounded by the ellipsoid and the plane. The graph is shown below.

[FIGURE 9 OMITTED]

Step 1. We pick three points [X.sub.0] = (0, 0, 0), [X.sub.1] = (1, -1, 0), and [X.sub.2] = (1, 0, -1) on the plane [P.sub.2] : x + y + z = 0 to, form the vectors u = [X.sub.1] - [X.sub.0] and v = [X.sub.2] - [X.sub.0]. We form orthonormal vectors

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (23)

Step 2. We compute X = A[X.sup.*], where [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], A = [[p.sub.1], [p.sub.2], n], [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Step 3. We substitute X with [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] into the formula F(x, y, z) = 0, or 4[x.sup.2] + 4[y.sup.2] + 4[z.sup.2] + 4xy + 4xz + 4yz - 5 = 0, to get an equation G(p, q, w) = 0.

Step 4. Solve G(p sin t cos s, [rho] sin t sins, [rho] cos t) = 0 for [rho], for which we obtained two values from Maple:

Case 1. If we use the following (see Maple worksheet, Section3-Example20.mws)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (24)

the volume of one portion of the region bounded by [P.sub.1] and F(x, y, z) = 0 is

[[integral].sup.2[pi].sub.0] [[integral].sup.[pi]/2.sub.0] ([[integral].sup.[rho](s,t).sub.0] ([[rho].sup.2] sin t) d[rho]) dtds = .4110463566,

when we use 200 points Simpson's Rule for approximation.

Case 2. If we use the, following

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (25)

the volume of the other portion of the region hounded by [P.sub.1] and F (x, y, z) = 0 is

[[integral].sup.2[pi].sub.0] [[integral].sup.[pi]/2.sub.0] ([[integral].sup.[rho](s,t).sub.0] ([[rho].sup.2] sin t) d[rho]) dtds = -7.867777159 (26)

when we use 200 points Simpson's Rule for approximation. Note that the total volume, by taking the absolute value, is about 8.278823557, which is accurate up to 8 decimal places. (See Maple worksheet .) It can be shown that the ellipsoid can be rotated to the form of

[x.sup.2]/[5/2] + [y.sup.2]/[5/2] + [z.sup.2]/[5/8] = 1,

which has a volume of 8.278823556.

An Exercise. We are given the surface S : [x.sup.2] + [y.sup.2]/4 + [z.sup.2]/9 = 1 and the plane P : x + y + z = [square root of 3]. Find the respective volumes bounded by the ellipsoid and the plane.

Answer: we get two values, 4.463982383 and -20.66875886, when we use n = 200 in Simpson's rule. (See Maple worksheet .) Readers can verify that the sum of the absolute values adds up to the volume of an ellipsoid by using the formula 4/3[pi]abc, when the ellipsoid is written as [x.sup.2]/[a.sup.2] + [y.sup.2]/[b.sup.2] + [z.sup.2]/[c.sup.2] = 1.

Remark 21 We note that the rotation method described here applies to the smooth surface satisfying F(x, y, z) = 0 and the method is valid only when [rho] is solvable by a CAS.

4 Conclusion

In an ordinary calculus textbook, the Divergence Theorem and Green's Theorem are usually taught in the last semester of a calculus sequence, to a group of students who completed some basic training in calculus. These theorems are important and yet difficult for many students. In this paper, we describe special ways of finding net signed areas and volumes through smooth parametric curves and surfaces.

The methods discussed in Sections 2 and 3 provide us with a way to find the net signed volume. The methods described here are accessible to those students who have knowledge in 2-D and 3-D Riemann integration. In particular, formulae (7) and (19) are natural extensions when we replace the graphs represented by y = f(x) and z = f(x, y) by a parametric curve or surface with respect to a line y = mx + b or a plane ax + by + cz = d, respectively. Consequently, the contents are accessible to a wider audience and can be used to prepare students for the study of the Divergence Theorem, Green's Theorem, and even Gauss-Bonnet Theorem in Differential Geometry.

Note that the results obtained in this paper are not possible without the help of technology. Dynamic Geometry collects the data, estimates the area, and allows us to construct the animation. The CAS aids us in the computation. graphing, and verification of conjectures. Authors anticipate that mathematical content will evolve as technological tools advance.

References

 Stewart. J. (2001), Calculus-Concepts and Contexts, second edition, Brooks/Cole Publishing, ISBN 0-534-37718-1.

 Larson. R-Hostetler, R., Edwards. B.. Essential Calculus-Early Transcendental Functions, Houghton Miffin Company, ISBN 13: 978-0-618-87918-2.

Software Packages

 [Maple] A product of Maplesoft, http://www.maplesoft.com/.

Supplemental Electronic Materials

 Yang. W.-C., a video clip which summarizes how ClassPad Manager is used in Section 1.1.

 Yang. W.-C.. ClassPad eActivity used in Section 1.1.

 Yang, W.-C., Lo, M.-L., Maple worksheet for Example 7.

 Yang, W.-C.. Lo, M.-L., Maple worksheet for Example 10.

 Yang, W.-C.. Lo, M.-L., Maple worksheet for Example 11.

 Yang. W.-C.. Lo, M.-L., Maple worksheet for Example 17.

 Yang, W.-C.. Lo, M.-L., Maple worksheet for Example 18.

 Yang, W.-C.. Lo. M.-L., Maple worksheet for Example 20.

 Yang, W.-C., Lo, M.-L., Maple worksheet for An Exercise in Section 3.

Wei-Chi Yang

Department of Mathematics and Statistics

USA

Min-Lin Lo

mlo@csusb.edu

Department of Mathematics

California State University San Bernardino

San Bernardino, CA 92407

USA
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Author: Printer friendly Cite/link Email Feedback Yang, Wei-Chi; Lo, Min-Lin Electronic Journal of Mathematics and Technology Report 1USA Jun 1, 2008 5667 Interactive geometry and critical points. Power series solutions of differential equations: programming and experimenting with technology. Area (Geometry) Calculus Calculus (Mathematics) Surfaces Volume (Cubic content) Volume (Cubic measurement)