Volume of revolution

rir0302

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Sep 11, 2019
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Screen Shot 2019-09-10 at 11.59.13 PM.png

Find the volume of the solid obtained by rotating the region A in the figure about x=−6.
Assume b=6, a=2.

Then, find the volume of the solid obtained by rotating the region 𝐵 in the figure about 𝑥=−3.
Assume b=4 and a=2.


I think I need help with the basic concept too because I'm stuck. For the first part, I tried doing the doing the disk formula for the equation, but I don't understand how to rotate around x=-6. Could someone show me the steps for the question? Thanks.
 

MarkFL

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Nov 24, 2012
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Hello, and welcome to FMH! :)

Let's let the axis of rotation be:

\(\displaystyle x=-k\) where \(0<k\)

Using the shell method, an element of the volume of the solid can be given as:

\(\displaystyle dv=2\pi(k+x)y\,dx=2\pi(x^3+kx^2+bx+bk)\,dx\)

Hence:

\(\displaystyle V=2\pi\int_0^a x^3+kx^2+bx+bk\,dx=2\pi\left[\frac{1}{4}x^4+\frac{k}{3}x^3+\frac{b}{2}x^2+bkx\right]_0^a=\frac{\pi a}{6}\left(3a^3+4a^2k+6ab+12bk\right)\)

Now we have a formula into which we can place the parameters \(a,\,b,\,k\).

See if you can derive a similar formula for region A.
 
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