Volumes of solids of revolution

a) $V=\pi {\int }_0^4(\sqrt{x}{)}^2\mathrm{d}x=\pi {\int }_0^{4}x\mathrm{d}x=\pi [\frac{x^2}{2}{]}_0^4=8\pi .$

b) $V=\pi {\int }_0^3{x}^2\mathrm{d}x=\pi [\frac{x^3}{3}{]}_0^3=9\pi .$

Check: the solid is a cone of radius $3$ and height $3$, with $V=\frac{1}{3}\pi \cdot 3^2\cdot 3=9\pi$.

Answer: a) $8\pi$; b) $9\pi$.

a) $x=x^2\Rightarrow x(1-x)=0$, so $x=0$ and $x=1$: the points are $(0,0)$ and $(1,1)$. Between them, $x\ge x^2$.

b) About the $x$-axis: outer radius $y=x$, inner radius $y=x^2$ (washers). $$V=\pi {\int }_0^1(x^2-x^4)\mathrm{d}x=\pi [\frac{x^3}{3}-\frac{x^5}{5}{]}_0^1=\pi (\frac{1}{3}-\frac{1}{5})=\frac{2\pi }{15}.$$

c) About the $y$-axis you must solve for $x$: the line is $x=y$ and the parabola is $x=\sqrt{y}$ (for $x\ge 0$). For each $y$ between $0$ and $1$, the curve further from the axis is $x=\sqrt{y}$ (outer), and the line $x=y$ is the inner one: $$V=\pi {\int }_0^1(y-y^2)\mathrm{d}y=\pi [\frac{y^2}{2}-\frac{y^3}{3}{]}_0^1=\frac{\pi }{6}.$$

Answer: a) $(0,0)$ and $(1,1)$; b) $\frac{2\pi }{15}$; c) $\frac{\pi }{6}$.

The curve cuts the $x$-axis at $x=1$, and at $x={\mathrm{e}}^2$ it has the value $y=2$. The region lies between $x=1$ and $x={\mathrm{e}}^2$.

a) About the $x$-axis: $V=\pi {\int }_1^{{\mathrm{e}}^2}(\ln x{)}^2\mathrm{d}x$. By parts twice ($u={\ln }^{2}x$, $\mathrm{d}v=\mathrm{d}x$): $$\int (\ln x{)}^2\mathrm{d}x=x(\ln x{)}^2-2x\ln x+2x+C.$$ Evaluating: at ${\mathrm{e}}^2$ it gives $4{\mathrm{e}}^2-4{\mathrm{e}}^2+2{\mathrm{e}}^2=2{\mathrm{e}}^2$, and at $1$ it gives $2$. Then $$V=\pi (2{\mathrm{e}}^2-2).$$

b) About the $y$-axis: solve $x={\mathrm{e}}^y$ for $y$ between $0$ and $2$. The region lies between the curve $x={\mathrm{e}}^y$ (left edge, inner) and the line $x={\mathrm{e}}^2$ (right edge, outer): $$V=\pi {\int }_0^2({\mathrm{e}}^4-{\mathrm{e}}^{2y})\mathrm{d}y=\pi [{\mathrm{e}}^{4}y-\frac{1}{2}{\mathrm{e}}^{2y}{]}_0^2=\pi (2{\mathrm{e}}^4-\frac{1}{2}{\mathrm{e}}^4+\frac{1}{2})=\frac{\pi (3{\mathrm{e}}^4+1)}{2}.$$

c) With the calculator, $V_a=2\pi ({\mathrm{e}}^2-1)\approx 40.1$ and $V_b\approx 259$.

Answer: a) $2\pi ({\mathrm{e}}^2-1)$; b) $\frac{\pi (3{\mathrm{e}}^4+1)}{2}$; c) $40.1$ and $259$.

In the exam: in b) what earns the marks is identifying the radii in $\mathrm{d}y$ and the limits $0$ and $2$ (not $1$ and ${\mathrm{e}}^2$). If you are given the integral to calculate with the calculator, write down the expression you have set up first.