Use both the washer method and the shell method to find the volume of the solid that is generated when the region
in the first quadrant bounded by , and is revolved about the line .
Set up the integral that gives the volume of the solid as a single integral if possible using the disk/washer method.
Select the correct choice below and fill in any answer boxes within your choice.
(Type exact answers.)
A.
B.

Ask by Garrett Garza. in the United States

Mar 29,2025

Question

Use both the washer method and the shell method to find the volume of the solid that is generated when the region
in the first quadrant bounded by , and is revolved about the line .
Set up the integral that gives the volume of the solid as a single integral if possible using the disk/washer method.
Select the correct choice below and fill in any answer boxes within your choice.
(Type exact answers.)
A.
B.

Ask by Garrett Garza. in the United States

Mar 29,2025

UpStudy AI · Accepted Answer

**Washer Method**

Since the solid is generated by revolving the region about the vertical line $$ x=2 $$, it is natural to take horizontal slices. In a horizontal slice at a fixed $$ y $$ (with $$ 0\le y\le 4 $$), the region extends in $$ x $$ from the line $$ x=0 $$ to the curve $$ y=x^2 $$, which can be solved for $$ x $$ as
$$
x=\sqrt{y}.
$$
When this horizontal segment is revolved about $$ x=2 $$ the distance from a point $$ x $$ to the line $$ x=2 $$ is $$ 2-x $$.

For a given $$ y $$:
- The **outer radius** corresponds to the leftmost point $$ x=0 $$ and is 
  $$
  R(y)=2-0=2.
  $$
- The **inner radius** corresponds to the rightmost point $$ x=\sqrt{y} $$ and is 
  $$
  r(y)=2-\sqrt{y}.
  $$

Thus, the area of the washer is
$$
A(y)=\pi\Bigl(R(y)^2-r(y)^2\Bigr)
=\pi\Bigl[4-(2-\sqrt{y})^2\Bigr].
$$

The volume using washers is then given by
$$
V=\pi\int_{y=0}^{4}\Bigl[4-(2-\sqrt{y})^2\Bigr]\,dy.
$$

**Shell Method**

Using vertical slices is also possible. A typical vertical slice at position $$ x $$ (with $$ 0\le x\le 2 $$ because the intersection of $$ y=x^2 $$ and $$ y=4 $$ occurs at $$ x=2 $$) has height determined by the difference between $$ y=4 $$ and $$ y=x^2 $$:
$$
\text{Height} = 4 - x^2.
$$
The distance from the slice at $$ x $$ to the line $$ x=2 $$ is the radius
$$
\text{Radius}=2-x.
$$

Thus, the volume using shells is
$$
V=2\pi\int_{x=0}^{2}(2-x)(4-x^2)\,dx.
$$

**Final Answers**

Washer Method (as a single integral):
$$
\boxed{V=\pi\int_{0}^{4}\Bigl[4-(2-\sqrt{y})^2\Bigr]\,dy.}
$$

Shell Method:
$$
\boxed{V=2\pi\int_{0}^{2}(2-x)(4-x^2)\,dx.}
$$
**Washer Method:** $$ V = \pi \int_{0}^{4} \left[4 - (2 - \sqrt{y})^2\right] \, dy $$ **Shell Method:** $$ V = 2\pi \int_{0}^{2} (2 - x)(4 - x^2) \, dx $$

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