Area and volumes between y=√x and y=x²

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Let $\Pi$ be the region between $y=\sqrt{x}$ and $y=x^2$. a) Find the intersections and the area of $\Pi$. b) Compute the volume of the solid obtained by rotating $\Pi$ about the $x$-axis. c) Compute the volume by rotating $\Pi$ about the $y$-axis using cylindrical shells $V=2\pi\int x[f(x)-g(x)]\,dx$. d) Compare the two volumes and explain the result using symmetry.

Solution

(a) Intersections: $\sqrt{x}=x^2\Rightarrow x=x^4\Rightarrow x(x^3-1)=0\Rightarrow x=0,\ x=1$, i.e. $(0,0)$ and $(1,1)$. On $[0,1]$ we have $\sqrt{x}\ge x^2$, so $A=\int_0^1(\sqrt{x}-x^2)\,dx=\left[\frac23 x^{3/2}-\frac{x^3}{3}\right]_0^1=\frac23-\frac13=\frac13.$

(b) Rotation about the $x$-axis (disks/washers). $V_x=\pi\int_0^1\!\left[(\sqrt{x})^2-(x^2)^2\right]dx=\pi\int_0^1(x-x^4)\,dx=\pi\left[\frac{x^2}{2}-\frac{x^5}{5}\right]_0^1=\frac{3\pi}{10}\approx0.9425.$

(c) Rotation about the $y$-axis (cylindrical shells). $V_y=2\pi\int_0^1 x(\sqrt{x}-x^2)\,dx=2\pi\int_0^1(x^{3/2}-x^3)\,dx=2\pi\left[\frac25 x^{5/2}-\frac{x^4}{4}\right]_0^1=2\pi\cdot\frac{3}{20}=\frac{3\pi}{10}\approx0.9425.$

(d) We obtain $V_x=V_y$. The region $\Pi$ is symmetric about the line $y=x$: swapping $x\leftrightarrow y$ turns $y=\sqrt{x}$ into $y=x^2$ and vice versa, so $\Pi$ is unchanged. Rotating about the $x$-axis or the $y$-axis therefore produces congruent solids of equal volume.

$\boxed{A=\frac13,\quad V_x=V_y=\frac{3\pi}{10}}$