Chapter 1: Q27E (page 1)

Question: To determine the integral in six different ways.

Short Answer

Expert verified

The integralhas been solved in six different ways.

Step by step solution

Given data

The region $E$ is the solid bounded by the surfaces $y = 4 - {x^2} - 4{z^2}$ and $y = 0$.

Concept used of multiple integral

A multiple integral is adefinite integralof afunction of several real variables, for instance,$f(x,y)$or$f(x,y,z)$.

Find the region for ${D_1}$

Let ${D_1},{D_2},{D_3}$ be the respective projections of $E$ on $xy,yz$ and $zx$-planes. The variable ${D_1}$ is the projection of $E$ on $xy$-plane. So set $z = 0$. Then, the equation becomes,

$\begin{array}{l}y = 4 - {x^2} - 4{z^2}\\y = 4 - {x^2} - 4{(0)^2}\\y = 4 - {x^2}\end{array}$

The graph of the above function is shown below in Figure 1.

From Figure 1 , it is observed that $x$ varies from $ - 2$ to 2 and $y$ varies from 0 to $4 - {x^2}$. To get the limits of $z$, Solve the given equations as below.

$\begin{array}{c}y = 4 - {x^2} - 4{z^2}4{z^2}\\ = 4 - {x^2} - y{z^2}\\ = \frac{{4 - {x^2} - y}}{4}z\\ = \pm \frac{1}{2}\sqrt {4 - {x^2} - y} {\rm{ }}\end{array}$

Hence, $E$

$ = \left\{ {(x,y,z)\mid - 2 \le x \le 2,0 \le y \le 4 - {x^2}, - \frac{1}{2}\sqrt {4 - {x^2} - y} \le z} \right.\left. { \le \frac{1}{2}\sqrt {4 - {x^2} - y} } \right\}$

Therefore, $\int_{ - 2}^2 {\int_0^{4 - {x^2}} {\int_{ - \frac{1}{2}\sqrt {4 - {x^2} - y} }^{\frac{1}{2}} f } } (x,y,z)dzdydx$

Also, from Figure 1 , it is observed that $y$ varies from 0 to 4, x varies from $ - \frac{1}{2}\sqrt {4 - y} $ to $\frac{1}{2}\sqrt {4 - y} $ and $z$ varies from $ - \frac{1}{2}\sqrt {4 - {x^2} - y} $ to $\frac{1}{2}\sqrt {4 - {x^2} - y} $.

Hence, $E$

$ = \left\{ {(x,y,z)\mid 0 \le y \le 4, - \frac{1}{2}\sqrt {4 - y} \le x \le \frac{1}{2}\sqrt {4 - y} , - \frac{1}{2}\sqrt {4 - {x^2} - y} \le z} \right.\left. { \le \frac{1}{2}\sqrt {4 - {x^2} - y} } \right\}$

Therefore, $\int_0^4 {\int_{ - \frac{1}{2}\sqrt {4 - y} - \frac{1}{2}\sqrt {4 - {x^2} - y} }^{\frac{1}{2}\sqrt {4 - y} } {\int_{\frac{1}{2}\sqrt {4 - {x^2} - y} } f } } (x,y,z)dzdxdy$

Find the region for ${D_2}$

The variable ${D_2}$ is the projection of $E$ on y z-plane . So set x=0. Then, the equation becomes,

$\begin{array}{l}y = 4 - {x^2} - 4{z^2}\\y = 4 - {(0)^2} - 4{z^2}\\y = 4 - 4{z^2}\end{array}$

The graph of the above function is shown below in Figure 2.

From Figure 2, it is observed that $y$ varies from 0 to 4 and $z$ varies from $ - \frac{1}{2}\sqrt {4 - y} $ to $\frac{1}{2}\sqrt {4 - y} $. To get the limits of $x$, Solve the given equations as below.

$\begin{array}{l}y = 4 - {x^2} - 4{z^2}\\{x^2} = 4 - 4{z^2} - y\\x = \pm \sqrt {4 - 4{z^2} - y} \end{array}$

Hence, $E$

$\quad = \left\{ {(x,y,z)\mid 0 \le y \le 4, - \frac{1}{2}\sqrt {4 - y} \le z \le \frac{1}{2}\sqrt {4 - y} , - \sqrt {4 - 4{z^2} - y} \le x \le } \right.\left. { - \sqrt {4 - 4{z^2} - y} } \right\}$

Also, from Figure 2 , it is observed that $z$ varies from $ - 1$ to 1, y varies from 0 to $4 - 4{z^2}$ and $x$ varies from $ - \sqrt {4 - 4{z^2} - y} $ to $\sqrt {4 - 4{z^2} - y} $.

Hence, $E$

$ = \left\{ {(x,y,z)\mid - 1 \le z \le 1,0 \le y \le 4 - 4{z^2}, - \sqrt {4 - y - 4{z^2}} \le x \le \sqrt {4 - y - 4{z^2}} } \right.$

Therefore, $E = \int_{ - 1}^1 {\int_0^{4 - 4{z^2}} {\int_{ - \sqrt {4 - y - 4{z^2}} }^{\sqrt {4 - y - 4{z^2}} } f } } (x,y,z)dxdydz$

Find the region for ${D_3}$

The variable ${D_3}$ is the projection of $E$ on z x-plane. So set y=0. Then, the equation becomes,

$\begin{array}{l}y = 4 - {x^2} - 4{z^2}\\(0) = 4 - {x^2} - 4{z^2}\\{x^2} = 4 - 4{z^2}\\{x^2} + 4{z^2} = 4\end{array}$

The graph of the above function is shown below in Figure 3 .

From Figure 3 , it is observed that $x$ varies from $ - 2$ to $2, z$ varies from $ - \frac{1}{2}\sqrt {4 - {x^2}} $ to $\frac{1}{2}\sqrt {4 - {x^2}} $ and $y$ varies from 0 to $y = 4 - {x^2} - 4{z^2}$.

Hence, $E$

$ = \left\{ {(x,y,z)\mid - 2 \le x \le 2, - \frac{1}{2}\sqrt {4 - {x^2}} \le z \le \frac{1}{2}\sqrt {4 - {x^2}} ,0 \le y \le 4 - {x^2}} \right.\left. { - 4{z^2}} \right\}$

Therefore, $E = \int_{ - 2}^2 {\int_{ - \frac{1}{2}\sqrt {4 - {x^2}} }^{\frac{1}{2}\sqrt {4 - {x^2}} } {\int_0^{4 - {x^2} - 4{z^2}} f } } (x,y,z)dydzdx$

Also, from Figure 3 , it is observed that $z$ varies from $ - 1$ to $1, x$ varies from $ - \sqrt {4 - 4{z^2}} $ to $\sqrt {4 - 4{z^2}} $ and $y$ varies from 0 to $y = 4 - {x^2} - 4{z^2}$.

Hence, $E$

$ = \left\{ {(x,y,z)\mid - 1 \le z \le 1, - \sqrt {4 - 4{z^2}} \le x \le \sqrt {4 - 4{z^2}} ,0 \le y \le 4 - {x^2} - 4{z^2}} \right\}$

Therefore, $E = \mathop \smallint \nolimits^1 \mathop \smallint \nolimits^{\sqrt {4 - 4{z^2}} } \mathop \smallint \nolimits^{4 - {x^2} - 4{z^2}} f(x,y,z)dydxdz$