Let $R$ be a region bounded by $y=x^2$, $y=2x$, and $y=3$. Solid $S$ is obtained by rotating region $R$ about $x=-1$. 
Let $R$ be a region bounded by $y=x^2$, $y=2x$, and $y=3$. Solid $S$ is obtained by rotating region $R$ about $x=-1$. UPDATE: the region R is the region bounded below the line y=3.


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*Write the integral for the volume of $S$ using the washer method.(no need to compute it)

*Find the volume of $S$ using the shell method.

Can someone help me with this problem please?
So for the first part of the problem, I am getting $$V = \pi  \int_0^3(\sqrt y - (-1)) dy$$. I'm not sure if I did it right.
For the second part, I got the radius is $x+1$ and the height is $3$.
Then $$V = 2\pi\int3(x+1)dx$$ (I'm not sure what the bounds are).
I am having trouble with this problem. Please help me out. Thank you. Also, I am not familiar with the proper tools on this site for math symbols and such. Sorry.
 A: There is some ambiguity here because the region $R$ is not clearly defined.  There are two possible enclosed regions, shown in the figure.  I have elected to use $R = B$ in the figure shown.  However, the solution for $R = A$ is substantially similar, with only minor modifications.

To see how we would set it up, we have to consider a representative washer for a corresponding $y$-coordinate.  We first observe that the region $R$ is given by $y \ge 3$, $y \le 2x$, and $y \ge x^2$.  Hence the outer radius of the washer is given by $r_o(y) = \sqrt{y} + 1$:  note that because $R$ is rotated about the axis $x = -1$ and not the $x$-axis, we must add $1$ the radii.
Similarly, the inner radius of the washer is $r_i(y) = y/2 + 1$.  Thus the differential volume of a representative washer is $$dV = \pi(r_o^2(y) - r_i^2(y)) \, dy = \pi\left(\left(\sqrt{y} + 1\right)^2 - (y/2 + 1)^2\right) \, dy.$$  The total volume is found by integrating over $y \in [3, 4]$, since $(2,4)$ is the common intersection point of $y = x^2$ and $y = 2x$ and is the upper bound for the $y$-interval that contains $R$.  Hence $$V = \int_{y=3}^4 \pi\left(\left(\sqrt{y} + 1\right)^2 - (y/2 + 1)^2\right) \, dy.$$
For the shell method, we integrate with respect to $x$.  This is a bit more complicated because while the interval of integration is $x \in [1.5, 2]$, the lower boundary of the shell's height is not continuous.  The upper boundary is of course $y = 2x$, but when $x \in [1.5, \sqrt{3}]$, the lower boundary is $y = 3$, and when $x \in (\sqrt{3}, 2]$, the lower boundary is $y = x^2$.  So for the first part, a representative shell has radius $r = x + 1$ (again, remembering that the axis of revolution is $x = -1$) and height $h_1 = 2x - 3$; thus the differential volume is $$dV_1 = 2\pi r h_1 \, dx = 2\pi(x+1)(2x-3) \, dx.$$  For the second part, the representative shell has again radius $r = x + 1$, but now the height function is $h_2 = 2x - x^2$, so $$dV_2 = 2\pi r h_2 \, dx = 2\pi(x+1)(2x-x^2) \, dx.$$  Thus our total volume is $$V = \int_{x=3/2}^{\sqrt{3}} 2\pi(x+1)(2x-3) \, dx + \int_{x=\sqrt{3}}^2 2\pi(x+1)(2x-x^2) \, dx.$$  You may perform the computations to verify both methods yield the same volume of $$\left(\frac{91}{12} - 4 \sqrt{3}\right)\pi.$$
