Part 2 of 2 My score: $$ 12 / 17 $$ pts (70.59\%) Save Let $$ R $$ be the region bounded by $$ y=x^{2}, x=1 $$, and $$ y=0 $$. Use the shell method to find the volume of the solid generated when $$ R $$ is revolved about the line $$ y=-1 $$. 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. $$ \int_{0}^{1}(2 \pi(y+1)(1-\sqrt{y})) d y $$ B. $$ \int() d x $$ The volume is $$ \square $$ (Type an exact answer.)

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We want the volume of the solid obtained by revolving the region
$$
R = \{(x,y) \mid 0 \le y \le x^2,\; x\le 1,\; y\ge 0\}
$$
about the line $$ y = -1 $$. Using the shell method with horizontal slices is natural when revolving about a horizontal line.

1. For a horizontal slice at a level $$ y $$ (where $$ 0 \le y \le 1 $$), the slice extends in the $$ x $$-direction from the curve $$ y=x^2 $$ to $$ x=1 $$. Solving $$ y=x^2 $$ for $$ x $$ gives $$ x=\sqrt{y} $$. Hence, the length (height) of the shell is
$$
h(y)= 1 - \sqrt{y}.
$$

2. The radius of the shell is the distance from the horizontal line $$ y = -1 $$ to the shell at height $$ y $$, which is
$$
r(y)= y - (-1)= y+1.
$$

3. The volume of a typical shell is
$$
dV = 2\pi \,( \text{radius} ) \,(\text{height})\,dy = 2\pi (y+1)(1-\sqrt{y})\,dy.
$$

4. Integrate from $$ y=0 $$ to $$ y=1 $$:
$$
V = 2\pi \int_{0}^{1} (y+1)(1-\sqrt{y})\,dy.
$$
Thus, the correct setup is choice A:
$$
\int_{0}^{1} \left(2\pi (y+1)(1-\sqrt{y})\right) \, dy.
$$

Now, we proceed to evaluate the integral.

First, expand the integrand:
$$
(y+1)(1-\sqrt{y}) = y + 1 - y\sqrt{y} - \sqrt{y} = y+1-y^{3/2}-y^{1/2}.
$$

So the volume becomes:
$$
V = 2\pi \int_{0}^{1} \left(1 + y - y^{1/2} - y^{3/2}\right) \, dy.
$$

Now, integrate term-by-term:
$$
\int_{0}^{1}1\,dy = 1,
$$
$$
\int_{0}^{1} y\,dy = \frac{1}{2},
$$
$$
\int_{0}^{1} y^{1/2}\,dy = \frac{2}{3},
$$
$$
\int_{0}^{1} y^{3/2}\,dy = \frac{2}{5}.
$$

Thus,
$$
\int_{0}^{1} \left(1 + y - y^{1/2} - y^{3/2}\right) \, dy = 1 + \frac{1}{2} - \frac{2}{3} - \frac{2}{5}.
$$

Finding a common denominator (which is 30) we have:
$$
1 = \frac{30}{30},\quad \frac{1}{2} = \frac{15}{30},\quad \frac{2}{3} = \frac{20}{30},\quad \frac{2}{5} = \frac{12}{30}.
$$
Thus,
$$
\frac{30}{30} + \frac{15}{30} - \frac{20}{30} - \frac{12}{30} = \frac{30+15-20-12}{30} = \frac{13}{30}.
$$

Substitute back into the expression for $$ V $$:
$$
V = 2\pi \cdot \frac{13}{30} = \frac{26\pi}{30} = \frac{13\pi}{15}.
$$

The final answers are as follows:

The integral setup is:
$$
2\pi \int_{0}^{1} (y+1)(1-\sqrt{y})\,dy.
$$

The volume of the solid is:
$$
\frac{13\pi}{15}.
$$
The volume of the solid is $$ \frac{13\pi}{15} $$.

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