Volumes of Solids of Revolution Method of Cylinders — Question 7

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Question 7

Find the volume of the solid obtained by rotating the region bounded by y=x,y=x2,y=\sqrt{x}, \qquad y=\frac{x}{2}, between their points of intersection, 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 6-4-007
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Question 7 – Solution

1. Find the intersections.

Both sides of x=x/2\sqrt{x}=x/2 are nonnegative. Squaring and factoring givesx=x24⇒x(x−4)=0⇒x=0,4.x=\frac{x^2}{4}\quad\Longrightarrow\quad x(x-4)=0\quad\Longrightarrow\quad x=0,4.Both values satisfy the original equation.

2. Identify radius and height.

Use vertical shells about the yy-axis. On [0,4][0,4], x≥x/2\sqrt{x}\ge x/2:r(x)=x,h(x)=x−x2.r(x)=x,\qquad h(x)=\sqrt{x}-\frac x2.

3. Set up and expand the shell integral.

V=2π∫04x(x−x2)dx=2π∫04(x3/2−x22)dx.V=2\pi\int_0^4x\left(\sqrt{x}-\frac x2\right)dx=2\pi\int_0^4\left(x^{3/2}-\frac{x^2}{2}\right)dx.

4. Integrate using the power rule.

∫x3/2dx=25x5/2,∫x22dx=x36.\int x^{3/2}\,dx=\frac25x^{5/2},\qquad\int\frac{x^2}{2}\,dx=\frac{x^3}{6}.V=2π[25x5/2−x36]04.V=2\pi\left[\frac25x^{5/2}-\frac{x^3}{6}\right]_0^4.

5. Apply the bounds and simplify.

Since 45/2=(4)5=324^{5/2}=(\sqrt4)^5=32,V=2π(645−646)=2π(192−16015)=64π15.V=2\pi\left(\frac{64}{5}-\frac{64}{6}\right)=2\pi\left(\frac{192-160}{15}\right)=\boxed{\frac{64\pi}{15}}.

All volumes are in cubic units.

Original worksheet page 2: question and worked solution for 6-4-007

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