Question

Volumes of Revolution Help.

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y = (1/x^2), y=0, x=3, x=5

about y=-1

I'm not sure what to do with this problem. I graphed it and it became more confusing once i saw it.

Answer 1

You are bounded by x=3 and x=5. Are you using the shell method, the disk method, or cross sections?

Answer 2

I would recommend the disk/washer method to find the volume. Draw a rectangle perpendicular to y=-1 and touching y=1/x*2.

Answer 3

Im not sure which method to use

Answer 4

Then your integral becomes:\[\int\limits_{3}^{5}\ [(\frac{ 1 }{x ^{2} }+1)^{2}-1^{2}]dx\]

Answer 5

Why is 1 added to (1/x^2)?

Answer 6

Because when you are revolving about the line y=a, the resulting radius is f(x)-a. In this case, a=-1, so the full radius from the line y=-1 to the function is f(x)-(1)

Answer 7

oops, that should be f(x)-(-1))

Answer 8

ah okay i see that

Answer 9

Now you get a much easier integral when the 1's cancel after squaring out the first part.

Answer 10

So, your integral becomes:\[\int\limits_{3}^{5}\left( \frac{ 1 }{ x ^{4} } +\frac{ 2 }{ x ^{2} }\right)dx\]

Answer 11

Now you can integrate and get:

\[-\frac{ 1 }{ 3 } x ^{-3}-2x ^{-1}\]

Evaluating at the given values, you have:

\[-\frac{ 1 }{ 3}(5)^{-3}-2(5)^{-1}-[-\frac{ 1 }{ 3 }3^{-3}-2(3)^{-1}\]

Answer 12

Finally, you can simplify that mess and get \[-\frac{ 1 }{375 }-\frac{ 2 }{ 5 }+\frac{ 1 }{ 81 }+\frac{ 2 }{ 3 }\]

Answer 13

I think you can see from here that you have to get an LCD to finish. Sorry, but not used to the equation editor so taking me awhile to type out replies.

Answer 14

By the way, I forgot the pi at the beginning, so don't forget to multiply your final answer by pi.

Answer 15

Thank you so much. You've been a huge help!

Answer 16

No prb.