Chapter 7.3, Problem 75E

To determine

To Find: The volume of a solid generated when R revolved about the y-axis.

Answer to Problem 75E

The volume of a solid is $128\prod$ which is option (B).

Explanation of Solution

Given information: The region R is enclosed by $y=x^2$ , $y=0$ and $x=4$ .

Calculation:

The region R is revolved around y- axis then, convert the equation in terms of y.

  $x=\sqrt{y}$

Now, find the boundaries for the integral. The upper and lower bound can be determined as:

The lower bound is $0$ as, it is given $y=0$ . The upper bound is:

  $\begin{array}{l}x=\sqrt{y}\\ 4=\sqrt{y}\\ y=16\end{array}$

Hence, the intersection point is $(4,16)$ so, the upper bound is $16$ .

According to the Washer method, area can be find out as:

  $\begin{array}{l}A=\prod (R^2-r^2)\\ R=4\\ r=\sqrt{y}\\ $ A=\prod (4^2-{(\sqrt{y})}^2)$A=\prod (16-y)\end{array}$

Now, find the volume:

  $\begin{array}{l}V={\displaystyle \underset{0}{\overset{16}{\int }}A(y)dy}\\ V=\prod {\displaystyle \underset{0}{\overset{16}{\int }}(16-y)dy}\\ V=\prod \left[{\displaystyle \underset{0}{\overset{16}{\int }}16dy-{\displaystyle \underset{0}{\overset{16}{\int }}ydy}}\right]\\ V=\prod \left[{(16)}^2-0-\frac{{(16)}^2}{2}-0\right]\\ V=\prod \left[256-128\right]\\ V=128\prod \end{array}$

Hence, volume of a solid is $128\prod$ .