Find the volume of the solid obtained by revolving about the y-axis the region enclosed by y = x² and x = y², using the disk/washer method.

Step-by-step solution

1. 1

   Solve for the intersection points. Substituting $x=y^2$ into $y=x^2$ gives $y=y^{4}$, so $y(y^{3}-1)=0$ and $y=0$ or $y=1$. The curves meet at $(0,0)$ and $(1,1)$, so the region sits inside the unit square.

2. 2

   Integrate in y, because the axis of revolution is the y-axis. Washers perpendicular to the $y$-axis are horizontal, so both boundaries must be written as $x$ in terms of $y$:

   $y=x^{2}\ \Rightarrow\ x=\sqrt{y}\ \ (\text{the right/outer boundary}),\qquad x=y^{2}\ \ (\text{the left/inner boundary})$

   On $0<y<1$ we have $\sqrt y>y^{2}$ (for example at $y=\tfrac14$: $0.5>0.0625$), confirming which is outer.

3. 3

   Set up the washer integral. The radius of a washer is the horizontal distance from the $y$-axis:

   $V=\pi\int_{0}^{1}\left[\left(\sqrt y\right)^{2}-\left(y^{2}\right)^{2}\right]dy=\pi\int_{0}^{1}\left(y-y^{4}\right)dy$

4. 4

   Evaluate.

   $V=\pi\left[\frac{y^{2}}{2}-\frac{y^{5}}{5}\right]_{0}^{1}=\pi\left(\frac12-\frac15\right)=\pi\cdot\frac{3}{10}$

   $\boxed{V=\dfrac{3\pi}{10}\approx 0.942}$

5. 5

   Cross-check with the shell method. Shells about the $y$-axis integrate in $x$ with radius $x$ and height $\sqrt x-x^{2}$:

   $V=2\pi\int_{0}^{1}x\left(\sqrt x-x^{2}\right)dx=2\pi\int_{0}^{1}\left(x^{3/2}-x^{3}\right)dx=2\pi\left(\frac25-\frac14\right)=2\pi\cdot\frac{3}{20}=\frac{3\pi}{10}$

   The two independent methods agree, which is the strongest available check on both the limits and the choice of outer/inner radius.