Let R be the region bound by sin(πx/12) and the x-axis. Use shells method to find the volume the solid obtained by revolving R about the line x = -1.

Question

So I think the graph should look like this:

And here is my formula and answer: $$\int_0^{12} 2\pi(1+x)\sin\left(\frac{π}{12}x\right)\ dx=336$$ I am not sure if my formula is correct. Thanks for your time!

Answer 1

Your answer is correct, The well-known equation

$$V=2\pi\int xy(x)~dx$$

is applicable when the axis of rotation is at the origin. Otherwise, it is easier to see what to do in terms of Pappus's $2^{nd}$ Centroid theorem: the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid $R$, i.e., $2πR$. Then $V=2\pi RA$. Now, the centroid distance from the $y$-axis is given by

$$R_x=\frac{1}{A}\int xy(x)~dx$$

But your axis of rotation is offset to $x=-1$, ergo

$$R_x=1+\frac{1}{A}\int xy(x)~dx$$

Thus, we must express the volume as

$$V=2\pi RA=2\pi \left(A+\int xy(x)~dx \right)=2\pi \int (1+x)y(x)~dx$$

just as you have shown. Your numerical result is easily verified as well.