FORM GR1268 Question 11

Question:
The region bounded by the curves \(y=x\) and \(y=x^2\) in the first quadrant of the \(xy\)-plane is rotated about the \(y\)-axis. The volume of the resulting solid of revolution is

(A) \(\displaystyle\frac{\pi}{12}\)     (B) \(\displaystyle\frac{\pi}{6}\)     (C) \(\displaystyle\frac{\pi}{3}\)     (D) \(\displaystyle\frac{2\pi}{3}\)     (E) \(\displaystyle\frac{3\pi}{2}\)

Answer:
(B)

Answer Key:
We will use the disc method for integration.
That is, we slice the solid perpendicular to the \(y\)-axis.

The cross sections of the solid are rings.
The diameter of the outer and inner circle depends on where on the \(y\)-axis you are at.
So, we want to express \(x\) in terms of \(y\):
The diameter of the inner circle is \(x=y\), derived from \(y=x\).
The diameter of the outer circle is \(x=\sqrt{y}\), derived from \(y=x^2\).

Thus, the area of the cross section at \(y\)-coordinate \(y\) is
\[A(y)=\pi(\sqrt{y}^2-y^2)=\pi(y-y^2).\]
To get the volume, integrate over \(0\le y\le 1\).
(Because the curves intersect at \((0,0)\) and \((1,1)\).)
\[\begin{align}
\int_0^1A(y)dy&=\int_0^1\pi(y-y^2)dy\\
&=\left.\pi\left(\frac{y^2}{2}-\frac{y^3}{3}\right)\right|_0^1\\
&=\pi\left(\frac{1}{2}-\frac{1}{3}\right)\\
&=\frac{\pi}{6}.
\end{align}\]