Volume generated by a solid revolution

Question

Find the volume of solid generated by region in the first quadrant bounded by curve $y=x^2$, below by the x axis and on the right by the line $x=2$ and about the line $x=-3$

How to set up integral in this question?

Please help.

Answer 1

First, shift the curve right such that the axis of revolution becomes the $y$ axis:

$$y = (x - 3)^2$$

Now put $x$ in terms of $y$ so that we can integrate along $y$:

$$x = \sqrt{y} + 3$$

Then, calculate the volume of the solid formed by revolving the region bounded on top by $y = 4$, below by the $x$ axis, on the left by the $y$ axis and about the $y$ axis:

$$R' = \pi \int_0^4 (\sqrt{y} + 3)^2 dy$$

Afterwards, subtract this volume off that of the cylinder formed by revolving the region bounded on top by $y = 4$, on the left by the $y$ axis, on the right by $x = 5$ and about the $y$ axis to get the required volume:

$$R = \pi\cdot(2 + 3)^2\cdot4 - R'$$

Answer 2

It will be easiest to use the Shell method, so we find the radius and height of the "shells." The radius (distance from axis of revolution to inside of shell) you can see is $3+x$ and the height is just the $y$ value ($=x^2$) so the volume is $$2\pi\int_0^2 x^2 (3+x)\text{d}x=24\pi$$