Volumes of Solids of Revolution Method of Cylinders — Question 10

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Question 10

Find the volume of the solid obtained by rotating the region bounded by y=ln⁡(1+x2)andy=0,y=\ln(1+x^2) \quad\text{and}\quad y=0, from x=0x=0 to x=1x=1, about the yy-axis.

See the diagram in the original worksheet below.

Rotate the shaded region about the yy-axis.

Original worksheet page 1: question and worked solution for 6-4-010
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Question 10 – Solution

1. Identify shell dimensions and set up the integral.

Use vertical shells about the yy-axis, with 0≤x≤10\le x\le1:r(x)=x,h(x)=ln⁡(1+x2).r(x)=x,\qquad h(x)=\ln(1+x^2).V=2π∫01xln⁡(1+x2)dx.V=2\pi\int_0^1x\ln(1+x^2)\,dx.

2. Substitute and change the bounds.

Let u=1+x2u=1+x^2. Then du=2xdxdu=2x\,dx, so xdx=du/2x\,dx=du/2. The bounds change as follows:x=0⇒u=1,x=1⇒u=2.x=0\Rightarrow u=1,\qquad x=1\Rightarrow u=2.

3. Rewrite the volume in terms of uu.

V=2π⋅12∫12ln⁡udu=π∫12ln⁡udu.V=2\pi\cdot\frac12\int_1^2\ln u\,du=\pi\int_1^2\ln u\,du.

4. Integrate the logarithm by parts.

Choose a=ln⁡ua=\ln u and db=dudb=du, giving da=du/uda=du/u and b=ub=u:∫ln⁡udu=uln⁡u−∫u1udu=uln⁡u−u.\int\ln u\,du=u\ln u-\int u\frac1u\,du=u\ln u-u.

5. Evaluate the endpoints.

V=π[ulnu−u]12=π[(2ln⁡2−2)−(ln⁡1−1)]=π(2ln⁡2−1)=2π(ln2−12).\begin{align*} V&=\pi\left[u\ln u-u\right]_1^2\\&=\pi\bigl[(2\ln2-2)-(\ln1-1)\bigr]\\&=\pi(2\ln2-1)=\boxed{2\pi\left(\ln2-\frac12\right)}. \end{align*}

All volumes are in cubic units.

Original worksheet page 2: question and worked solution for 6-4-010

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