Volumes of Solids of Revolution Method of Cylinders — Question 4

Question 4 – Solution

1. Identify shell radius and height.

Use vertical slices about the $y$-axis, with $0\le x\le e-1$:$$r(x)=x,\qquad h(x)=\ln(1+x)-0.$$

2. Set up the volume.

$$V=2\pi\int_0^{e-1}x\ln(1+x)\,dx.$$

3. Apply integration by parts.

Choose $u=\ln(1+x)$ and $dv=x\,dx$. Then $du=dx/(1+x)$ and $v=x^2/2$:$$\int x\ln(1+x)\,dx=\frac{x^2}{2}\ln(1+x)-\frac12\int\frac{x^2}{1+x}\,dx.$$

4. Divide and integrate the remaining terms.

Since $x^2=(x+1)(x-1)+1$,$$\frac{x^2}{1+x}=x-1+\frac1{1+x}.$$An antiderivative is therefore$$F(x)=\frac{x^2}{2}\ln(1+x)-\frac{x^2}{4}+\frac x2-\frac12\ln(1+x).$$

5. Apply the bounds and multiply by $2\pi$.

Using $\ln e=1$ and $F(0)=0$,$$\begin{align*} F(e-1)-F(0)&=\frac{(e-1)^2}{4}+\frac{e-1}{2}-\frac12=\frac{e^2-3}{4},\\V&=2\pi\cdot\frac{e^2-3}{4}=\boxed{\frac\pi2(e^2-3)}. \end{align*}$$

All volumes are in cubic units.