My tutor gave me this question and I don't know how to tackle this type of question.

Consider the solid of revolution obtained when the region surrounded by $x=2$, $x=3$, $y=1$ and $y=\frac{x}{x+3}$ is rotated about the line $y=1$. Find an expression for the cross-sectional area of this solid at a point $x$.

  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here and here. $\endgroup$ – Martin Sleziak Aug 15 '12 at 10:38

Here’s a rough sketch of the region, which I’ve labelled $R$. enter image description here

When you revolve the red line about the line $y=1$, it traces out a disk; this disk is one of the cross-sections of the resulting solid of revolution. The radius of the disk is the length of the red line, which is the $y$-coordinate at the top of the line minus the $y$-coordinate at the bottom. The $y$-coordinate at the top is $1$; the one at the bottom is $\frac{x}{x+3}$. Thus, the radius of the disk is


and the area of the disk is $$\pi\left(\frac3{x+3}\right)^2=\frac{9\pi}{(x+3)^2}\;.$$


If you can imagine the shape formed, you will see that the radius of the cross-section is simply the difference between $y=1$ and $y=x/(x+3)$, or $r = 1 - x/(x+3)$.

The area is therefore $\pi r^2 = \pi \left(1-\frac{x}{x+3}\right)^2$


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