Volumes of Solids of Revolution Method of Rings — Question 3

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Question 3

Find the volume of the solid obtained by rotating the region bounded by y=x2andy=1,y=x^2 \quad\text{and}\quad y=1, from x=−1x=-1 to x=1x=1, about the xx-axis.

See the diagram in the original worksheet below.

Rotate the shaded region about the xx-axis.

Original worksheet page 1: question and worked solution for 6-3-003
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Question 3 – Solution

1. Find the bounds and choose slices.

The intersections satisfy x2=1x^2=1, so x=−1x=-1 and x=1x=1. Use vertical slices of thickness dxdx because the rotation axis is horizontal.

2. Identify outer and inner radii.

On [−1,1][-1,1], 0≤x2≤10\le x^2\le1. Measured from the xx-axis,R(x)=1,r(x)=x2.R(x)=1,\qquad r(x)=x^2.The missing inner disk must be subtracted from the outer disk.

3. Set up the washer integral.

A(x)=π(R(x)2−r(x)2)=π(12−(x2)2)=π(1−x4).A(x)=\pi\bigl(R(x)^2-r(x)^2\bigr)=\pi\bigl(1^2-(x^2)^2\bigr)=\pi(1-x^4).V=π∫−11(1−x4)dx.V=\pi\int_{-1}^1(1-x^4)\,dx.

4. Integrate term by term.

∫1dx=x,∫x4dx=x55.\int1\,dx=x,\qquad\int x^4\,dx=\frac{x^5}{5}.V=π[x−x55]−11.V=\pi\left[x-\frac{x^5}{5}\right]_{-1}^1.

5. Evaluate both endpoints.

V=π[(1−15)−(−1+15)]=π(45+45)=8π5.\begin{align*} V&=\pi\left[\left(1-\frac15\right)-\left(-1+\frac15\right)\right]\\&=\pi\left(\frac45+\frac45\right)=\boxed{\frac{8\pi}{5}}. \end{align*}

All volumes are in cubic units.

Original worksheet page 2: question and worked solution for 6-3-003

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