Find the volume of the solid generated by rotating

Find the volume of the solid generated by rotating the region bounded by the curves $y = x^2$ and $y^2 = x$ about: 1° the $x$-axis, 2° the line $x = y$.

We handle the $x$-axis rotation directly, then use a symmetry argument for the trickier rotation about the diagonal line $x=y$.

Step 1: Find intersection points

From $y=x^2$ and $y^2=x$: $(x^2)^2=x \implies x^4-x=0 \implies x(x^3-1)=0$, so $x=0$ or $x=1$, giving points $(0,0)$ and $(1,1)$.

Step 2: Determine which curve is above

At $x=\frac14$: $y=x^2=\frac{1}{16}$ but $y=\sqrt x=\frac12$. So $y=\sqrt x$ (from $y^2=x$) lies above $y=x^2$ on $(0,1)$.

Part 1: Rotation about the $x$-axis

Washer method: outer radius $\sqrt x$, inner radius $x^2$:

\[V_x = \pi\int_0^1(x-x^4)\,dx = \pi\left[\frac{x^2}{2}-\frac{x^5}{5}\right]_0^1 = \pi\left(\frac12-\frac15\right) = \frac{3\pi}{10}\]

\[\boxed{V_x = \frac{3\pi}{10} \text{ cubic units}}\]

Part 2: Rotation about the line $x=y$

Step 1: A key symmetry

Swapping $x$ and $y$ turns $y=x^2$ into $x=y^2$ (i.e. $y^2=x$) and vice versa — exactly the two boundary curves swap roles. Since the line $x=y$ is precisely the fixed line of this swap, the line $x=y$ splits the region into two mirror-image halves: $R_1=\{x^2\le y\le x\}$ (below the line) and $R_2=\{x\le y\le\sqrt x\}$ (above the line), each the reflection of the other across $x=y$.

Step 2: Why only one half needs to be revolved

Revolving a planar region about an axis that runs through its interior, splitting it into two mirror-image halves, produces exactly the same solid as revolving just one of the halves alone: reflecting $R_1$ across the axis of revolution reproduces $R_2$, and a shape together with its own mirror image (across the axis of revolution) sweep out identical solids. So the true volume equals the volume swept by $R_1$ alone (not the sum of both halves).

Step 3: Apply Pappus's theorem to $R_1$

$R_1 = \{(x,y): 0\le x\le1,\ x^2\le y\le x\}$ lies entirely on the $x\ge y$ side of the line $x=y$, with perpendicular distance $d=\dfrac{x-y}{\sqrt2}$. By Pappus's theorem, $V=2\pi\bar d\,A = 2\pi\displaystyle\iint_{R_1}\frac{x-y}{\sqrt2}\,dA$.

Step 4: Compute the moment integral

\[\iint_{R_1}(x-y)\,dA = \int_0^1\int_{x^2}^{x}(x-y)\,dy\,dx = \int_0^1\left[\frac{x^2}{2}-x^3+\frac{x^4}{2}\right]dx\]

\[= \left[\frac{x^3}{6}-\frac{x^4}{4}+\frac{x^5}{10}\right]_0^1 = \frac16-\frac14+\frac{1}{10} = \frac{10-15+6}{60} = \frac{1}{60}\]

Step 5: Calculate the volume

\[V_{x=y} = 2\pi\cdot\frac{1}{\sqrt2}\cdot\frac{1}{60} = \frac{2\pi}{60\sqrt2} = \frac{\pi\sqrt2}{60}\]

\[\boxed{V_{x=y} = \frac{\pi\sqrt2}{60} \text{ cubic units}}\]

Step 6: Numerical value

\[V_{x=y} = \frac{\pi\sqrt2}{60} \approx 0.0740 \text{ cubic units}\]

Summary

\[\boxed{V_x = \frac{3\pi}{10}\approx0.942 \text{ cubic units}, \qquad V_{x=y} = \frac{\pi\sqrt2}{60}\approx0.0740 \text{ cubic units}}\]

Key insight: Rotating a region about a line of symmetry running through its own interior is a subtle case: the naive approach of integrating the full region's perpendicular extent on both sides double-counts, since the two mirror halves sweep out the identical solid. Recognizing the $x\leftrightarrow y$ symmetry between $y=x^2$ and $y^2=x$ is what makes the second rotation tractable at all.