13. Volume

Recall:   The Volume of a Solid computed using Thin Disks is V=abπf(x)2dxorV=abπf(y)2dy V=\int_a^b \pi f(x)^2\,dx \qquad \text{or} \qquad V=\int_a^b \pi f(y)^2\,dy

c1b. Volume using Thin Disks - Example 2

Now we come to volumes of revolution which do not have a classical formula:

The region between the curve x=1yx=\dfrac{1}{y} and the yy-axis for 1y41 \le y \le 4 is rotated about the yy-axis. Find the volume swept out.

eg_1_y_about_y_solid_anim

The curve is a function of yy; so this is a yy-integral and we cut up the yy-axis. Here is the region and one rotated disk:

The radius of the disk is r=x=1yr=x=\dfrac{1}{y}. So the area is A=πr2=πy2A=\pi r^2=\dfrac{\pi}{y^2} and the volume is V=14πy2dy=[πy]14=π4+π=3π4 V=\int_1^4 \dfrac{\pi}{y^2}\,dy =\left[-\,\dfrac{\pi}{y}\right]_1^4 =-\,\dfrac{\pi}{4}+\pi=\dfrac{3\pi}{4}

eg_1_y_about_y_disk_anim

Find the volume of the solid of revolution obtained by rotating the region under the graph of y=x1y=\sqrt{x-1} for x5x \le 5 about the xx-axis.

ex_parab_about_x_solid_anim

Answer

V=8πV=8\pi

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Solution

Since yy is a function of xx, we chop up the xx-axis, draw a vertical Riemann sum rectangle and see it rotates into a disk as shown.

With a=1a=1, b=5b=5 and r=y=x1r=y=\sqrt{x-1}, the volume is V=π15x12dx=π15(x1)dx=π[x22x]15=8π\begin{aligned} V&=\pi\int_1^5 \sqrt{x-1}^2\,dx =\pi\int_1^5 (x-1)\,dx \\ &=\pi\left[\dfrac{x^2}{2}-x\right]_1^5 =8\pi \end{aligned}

ex_parab_about_x_disk_anim
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