Let R be the region bounded by $$ \mathrm{y}=\mathrm{x}^{2}, \mathrm{x}=1 $$, and $$ \mathrm{y}=0 $$. Use the shell method to find the volume of the solid generated when R is revolved about the line $$ \mathrm{y}=-1 $$. 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.

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UpStudy AI · Accepted Answer

$$
V=2\pi\int_{0}^{1}(y+1)\Bigl(1-\sqrt{y}\Bigr)dy
$$

To explain briefly:

1. We choose horizontal shells (with thickness $$dy$$) since the axis of rotation $$y=-1$$ is horizontal.
2. For a horizontal line at height $$y$$ (with $$0\le y\le 1$$, because the maximum $$y$$-value in $$R$$ occurs at $$x=1$$ where $$y=1$$), the distance from the shell to the axis is the radius
   $$
   r=y-(-1)=y+1.
   $$
3. For a fixed $$y$$, the region extends in $$x$$ from the curve $$y=x^2$$ (i.e. $$x=\sqrt{y}$$) to the vertical line $$x=1$$. Therefore, the length (height of the shell) is
   $$
   1-\sqrt{y}.
   $$
4. Multiplying the circumference $$2\pi r$$, the height, and the thickness $$dy$$ gives the volume of a thin shell.
5. Integrating from $$y=0$$ to $$y=1$$ gives the total volume.

Thus, the integral is

$$
V=2\pi\int_{0}^{1}(y+1)(1-\sqrt{y})\,dy.
$$
$$ V=2\pi\int_{0}^{1}(y+1)(1-\sqrt{y})\,dy $$

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