Volumes of Solids of Revolution Method of Cylinders — Question 6

Question 6 – Solution

1. Determine the upper curve.

Let $g(x)=x-1-\ln x$. On $[1,e]$, $g(1)=0$ and $g'(x)=1-1/x\ge0$. Hence $x-1\ge\ln x$.

2. Identify shell dimensions and set up the volume.

Use vertical slices parallel to the $y$-axis:$$r(x)=x,\qquad h(x)=x-1-\ln x.$$$$V=2\pi\int_1^e x(x-1-\ln x)\,dx=2\pi\int_1^e(x^2-x-x\ln x)\,dx.$$

3. Integrate the logarithmic term by parts.

Let $u=\ln x$, $dv=x\,dx$, so $du=dx/x$ and $v=x^2/2$:$$\int x\ln x\,dx=\frac{x^2}{2}\ln x-\frac12\int x\,dx=\frac{x^2}{2}\ln x-\frac{x^2}{4}.$$

4. Combine the antiderivatives.

$$F(x)=\frac{x^3}{3}-\frac{x^2}{2}-\left(\frac{x^2}{2}\ln x-\frac{x^2}{4}\right)=\frac{x^3}{3}-\frac{x^2}{4}-\frac{x^2}{2}\ln x.$$

5. Apply the bounds.

$$F(e)=\frac{e^3}{3}-\frac{3e^2}{4},\qquad F(1)=\frac13-\frac14=\frac1{12}.$$$$V=2\pi\bigl(F(e)-F(1)\bigr)=\boxed{\pi\left(\frac{2e^3}{3}-\frac{3e^2}{2}-\frac16\right)}.$$

All volumes are in cubic units.