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Found in: Page 806

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Show the parametric equation$$x = a\sin u\cos v,y = b\sin u\sin v,z = c\cos u,0 \le u \le \pi ,0 \le v \le 2\pi$$ represents an ellipsoid.Draw the graph of an ellipsoid with the parametric equations$$x = a\sin u\cos v,y = b\sin u\sin v,z = c\cos u,0 \le u \le \pi ,0 \le v \le 2\pi {\rm{ for }}a = 1,b = 2,{\rm{ and }}c = 3$$Find the expression for surface area of the ellipsoid with the parametric equations$$x = a\sin u\cos v,y = b\sin u\sin v,z = c\cos u,0 \le u \le \pi ,0 \le v \le 2\pi {\rm{ for }}a = 1,b = 2,{\rm{ and }}c = 3$$

1. The parametric equations$$x = a\sin u\cos v,y = b\sin u\sin v,z = c\cos u,0 \le u \le \pi ,0 \le v \le 2\pi$$ are represented an ellipsoid.
2. The answer is given below
3. The expression for surface area of the ellipsoid with the parametric equations$$x = a\sin u\cos v,y = b\sin u\sin v,z = c\cos u,0 \le u \le \pi ,0 \le v \le 2\pi {\rm{ for }}a = 1,b = 2{\rm{, and }}c = 3$$is$$A(S) = \int_0^{2\pi } {\int_0^\pi {\sqrt {36{{\sin }^4}u{{\cos }^2}v + 9{{\sin }^4}u{{\sin }^2}v + 4{{\sin }^2}u{{\cos }^2}u} } } dudv$$.
See the step by step solution

## Step 1: Expression for equation of ellipsoid

(a)

Write the expression for equation of ellipsoid:

$$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$$ …… (1)

## Step 2: Use the equation of ellipsoid for further calculation

Write the parametric equation as follows,

$$x = a\sin u\cos v,y = b\sin u\sin v,z = c\cos u,0 \le u \le \pi ,0 \le v \le 2\pi$$

Rewrite the parametric equation as follows,

$$\frac{x}{a} = \sin u\cos v,\frac{y}{b} = \sin u\sin v,\frac{z}{c} = \cos u$$

Take square and add the expression as follows,

$${\left( {\frac{x}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} + {\left( {\frac{z}{c}} \right)^2} = {(\sin u\cos v)^2} + {(\sin u\sin v)^2} + {(\cos u)^2}$$

Simplify the expression as follows,

\begin{align} & frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}+\frac{{{z}^{2}}}{{{c}^{2}}}={{\sin }^{2}}u{{\cos }^{2}}v+{{\sin }^{2}}u{{\sin }^{2}}v+{{\cos }^{2}}u \\ & ={{\sin }^{2}}u\left( {{\cos }^{2}}v+{{\sin }^{2}}v \right)+{{\cos }^{2}}u \\ & ={{\sin }^{2}}u(1)+{{\cos }^{2}}u\left( \because {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \right) \\ & ={{\sin }^{2}}u+{{\cos }^{2}}u \\ \end{align}

Then,

$$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$$ …… (2)

As the equations (1) and (2) are similar, the parametric equation represents an ellipsoid.

Thus, the parametric equations$$x = a\sin u\cos v,y = b\sin u\sin v,z = c\cos u,0 \le u \le \pi ,0 \le v \le 2\pi$$ are represented an ellipsoid.

## Step 3: Consider the values of scalar parameters in the parametric equations

(b)

Consider the values of scalar parameters in the parametric equations as follows,

$$a = 1,b = 2,{\rm{ and }}c = 3$$

Substitute $$1{\rm{ for }}a,2{\rm{ for }}b,{\rm{ and }}3{\rm{ for }}c$$ in the parametric equations and rewrite the parametric equations of the ellipsoid as follows.

$$\begin{array}{l}x = (1)\sin u\cos v,y = (2)\sin u\sin v,z = (3)\cos u,0 \le u \le \pi ,0 \le v\\ \le 2\pi \\x = \sin u\cos v,y = 2\sin u\sin v,z = 3\cos u,0 \le u \le \pi ,0 \le v \le 2\pi \end{array}$$ …… (3)

From the analysis in part (a) and equation (3), draw the graph of an ellipsoid with the parametric equations $$x = a\sin u\cos v,y = b\sin u\sin v,z = c\cos u,0 \le u \le \pi ,0 \le v \le 2\pi {\rm{ for }}a = 1,b = 2,{\rm{ and }}c = 3$$ as shown in Figure 1.

Thus, the graph of an ellipsoid of the parametric equations $$x = a\sin u\cos v,y = b\sin u\sin v,z = c\cos u,0 \le u \le \pi ,0 \le v \le 2\pi {\rm{ for }}a = 1,b = 2,{\rm{ and }}c = 3$$ is drawn.

## Step 4: Expression to find the surface area of the plane

(c)

Write the expression to find the surface area of the plane with vector equation$${\rm{r}}(u,v)$$

$$A(S)=\iint_{D}{\left| {{\text{r}}_{u}}\times {{\text{r}}_{v}} \right|}dA$$ …… (4)

Here,

$${r_u}$$ Is the derivative of vector equation $${\bf{r}}(u,v)$$ with respect to the parameter $$u$$ and

$${{\rm{r}}_v}$$ Is the derivative vector equation $${\rm{r}}(u,v)$$ with respect to the parameter $$v{\rm{. }}$$

Write the expression to find $${r_u}.$$

$${{\rm{I}}_u} = \frac{\partial }{{\partial u}}(\;{\rm{T}}(u,v))$$ …… (5)

Write the expression to find $${{\rm{r}}_v}.$$

$${\Gamma _v} = \frac{d}{{{\partial _v}}}(\Gamma (u,v))$$ ...... (6)

## Step 5: Use the equations (4), (5), and (6) for further calculation

Write the vector function $${\rm{r}}(u,v)$$ for the parametric equations in equation (3),

$${\rm{r}}(u,v) = \langle \sin u\cos v,2\sin u\sin v,3\cos u\rangle ,0 \le u \le \pi ,0 \le v \le 2\pi$$

Calculation of$$ru$$:

Substitute $$\langle \sin u\cos v,2\sin u\sin v,3\cos u\rangle {\rm{ for r}}(u,v)$$ in equation (5),

$$\begin{array}{l}{{\rm{r}}_u} = \frac{\partial }{{\partial u}}\langle \sin u\cos v,2\sin u\sin v,3\cos u\rangle \\ = \left\langle {\frac{\partial }{{\partial u}}(\sin u\cos v),\frac{\partial }{{\partial u}}(2\sin u\sin v),\frac{\partial }{{\partial u}}(3\cos u)} \right\rangle \\ = \left\langle {\cos v\frac{\partial }{{\partial u}}(\sin u),2\sin v\frac{\partial }{{\partial u}}(\sin u),3\frac{\partial }{{\partial u}}(\cos u)} \right\rangle \\ = \langle \cos u\cos v,2\cos u\sin v, - 3\sin u\rangle \end{array}$$

Calculation of$$rv$$:

Substitute $$\langle \sin u\cos v,2\sin u\sin v,3\cos u\rangle {\rm{ for r}}(u,v)$$ in equation (6),

Calculation of$$ru \times rv$$:

Substitute $$\langle \cos u\cos v,2\cos u\sin v, - 3\sin u\rangle {\rm{ for }}{{\rm{r}}_u}{\rm{ and }}\langle - \sin u\sin v,2\sin u\cos v,0\rangle {\rm{ for }}{{\rm{r}}_t}$$ in the expression$$ru \times rv$$,

$${{\rm{r}}_u} \times {{\rm{r}}_v} = \left\langle {\begin{array}{*{20}{l}}{\cos u\cos v,}\\{2\cos u\sin v, - 3\sin u}\end{array}} \right\rangle \times \left\langle {\begin{array}{*{20}{l}}{ - \sin u\sin v}\\{2\sin u\cos v,0}\end{array}} \right\rangle$$

Rewrite and compute the expression as follows,

$$\begin{array}{l}{{\rm{r}}_u} \times {{\rm{r}}_v} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{\cos u\cos v}&{2\cos u\sin v}&{ - 3\sin u}\\{ - \sin u\sin v}&{2\sin u\cos v}&0\end{array}} \right|\\ = \left( {\begin{array}{*{20}{c}}{\left| {\begin{array}{*{20}{c}}{2\cos u\sin v}&{ - 3\sin u}\\{2\sin u\cos v}&0\end{array}} \right|{\rm{i}} - \left| {\begin{array}{*{20}{c}}{\cos u\cos v}&{ - 3\sin u}\\{ - \sin u\sin v}&0\end{array}} \right|{\rm{j}}}\\{ + \left| {\begin{array}{*{20}{c}}{\cos u\cos v}&{2\cos u\sin v}\\{ - \sin u\sin v}&{2\sin u\cos v}\end{array}} \right|{\rm{k}}}\end{array}} \right)\\ = \left\{ {\begin{array}{*{20}{l}}{\left( {0 - \left( {6{{\sin }^2}u\cos v} \right)} \right){\rm{i}} - \left( {0 - 3{{\sin }^2}u\sin v} \right){\rm{j}}}\\{ + \left( {2\sin u\cos u\cos 2v - \left( { - 2\sin u\cos u{{\sin }^2}v} \right)} \right){\rm{k}}}\end{array}} \right\}\\ = - 6{\sin ^2}u\cos v{\rm{i}} + 3{\sin ^2}u\sin v{\rm{j}} + 2\sin u\cos u\left( {{{\cos }^2}v + {{\sin }^2}v} \right){\rm{k}}\end{array}$$

Simplify the expression as follows,

$${{\rm{r}}_u} \times {{\rm{r}}_v} = \left\langle { - 6{{\sin }^2}u\cos v,3{{\sin }^2}u\sin v,2\sin u\cos u} \right\rangle$$

Substitute$$\left\langle { - 6{{\sin }^2}u\cos v,3{{\sin }^2}u\sin v,2\sin u\cos u} \right\rangle {\rm{ for }}{{\rm{r}}_u} \times {{\rm{r}}_v}$$ in equation (4),

Apply the limits$$u,v$$ and rewrite the expression as follows,

$$A(S) = \int_0^{2\pi } {\int_0^\pi {\sqrt {36{{\sin }^4}u{{\cos }^2}v + 9{{\sin }^4}u{{\sin }^2}v + 4{{\sin }^2}u{{\cos }^2}u} } } dudv$$

Thus, the expression for surface area of the ellipsoid with the parametric equations$$x = a\sin u\cos v,y = b\sin u\sin v,z = c\cos u,0 \le u \le \pi ,0 \le v \le 2\pi {\rm{ for }}a = 1,b = 2{\rm{, and }}c = 3$$is$$A(S) = \int_0^{2\pi } {\int_0^\pi {\sqrt {36{{\sin }^4}u{{\cos }^2}v + 9{{\sin }^4}u{{\sin }^2}v + 4{{\sin }^2}u{{\cos }^2}u} } } dudv$$.