Part 2 of 2
Let R be the region bounded by , and . Use the shell method to find the volume of the solid
generated when R is revolved about the line . 10 of 17 .
Set up the integral that gives the volume of the solid using the shell method. Use increasing limits of integration.
Select the correct choice below and fill in the answer boxes to complete your choice.
(Type exact answers.)
A. dy
B.
The volume of the solid is
(Type an exact answer.)

Ask by Barrett Mills. in the United States

Mar 29,2025

Question

Part 2 of 2
Let R be the region bounded by , and . Use the shell method to find the volume of the solid
generated when R is revolved about the line . 10 of 17 .
Set up the integral that gives the volume of the solid using the shell method. Use increasing limits of integration.
Select the correct choice below and fill in the answer boxes to complete your choice.
(Type exact answers.)
A. dy
B.
The volume of the solid is
(Type an exact answer.)

Ask by Barrett Mills. in the United States

Mar 29,2025

UpStudy AI · Accepted Answer

Let the region be  
$$
R=\{(x,y)\mid 0\le x\le1,\;0\le y\le x^2 \}.
$$
When revolving $$R$$ about the vertical line $$x=-8$$, using the shell method with vertical slices (i.e. integrating with respect to $$x$$):

1. The radius of a typical shell is the horizontal distance from $$x$$ to $$-8$$:  
   $$
   \text{radius}=x-(-8)=x+8.
   $$
2. The height of the shell is the value of the function:  
   $$
   \text{height}=x^2.
   $$
3. The differential thickness is $$dx$$.

Thus, the volume element is  
$$
dV=2\pi\;(\text{radius})\;(\text{height})\;dx=2\pi(x+8)x^2\,dx.
$$

The integral giving the volume is  
$$
\int_{x=0}^{1}2\pi(x+8)x^{2}\,dx.
$$

Evaluating the integral:
$$
\begin{aligned}
V &= 2\pi \int_{0}^{1}(x+8)x^{2}\,dx \
  &= 2\pi \int_{0}^{1}(x^3+8x^2)\,dx \
  &= 2\pi\left[\int_{0}^{1}x^3\,dx +8\int_{0}^{1}x^2\,dx\right] \
  &= 2\pi\left[\frac{x^4}{4}\Big|_0^1+8\cdot\frac{x^3}{3}\Big|_0^1\right] \
  &= 2\pi\left[\frac{1}{4}+8\cdot\frac{1}{3}\right] \
  &= 2\pi\left[\frac{1}{4}+\frac{8}{3}\right] \
  &= 2\pi\left[\frac{3+32}{12}\right] \
  &= 2\pi\left(\frac{35}{12}\right) \
  &= \frac{35\pi}{6}.
\end{aligned}
$$

Thus, the correct choice is B with the integral  
$$
\int_{0}^{1} 2\pi(x+8)x^2\,dx,
$$
and the volume of the solid is  
$$
\frac{35\pi}{6}.
$$
The volume of the solid is $$ \frac{35\pi}{6} $$.

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