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November 27, 2014

November 27, 2014

Posted by **Mooch** on Tuesday, March 13, 2012 at 4:49pm.

Here is what I have so far:

Radius = 1/(x^3)

Area of Cross Section = pi(1/(x^3))^2

Simplified: pi(1/(x^6))

Volume = (definite integral from 1 to 3) pi(1/(x^6)) dx

= pi( -1 / 5(3)^5) - pi(-1 / 5(1)^5)

= pi (-1 / 1215) - pi (-1 / 5)

= pi(242 / 1215) = 0.625732858

Is this right?

Sincerely,

Mooch

- Calculus Check -
**Steve**, Tuesday, March 13, 2012 at 5:00pmjust right. good job.

- Calculus Check -
**Mooch**, Tuesday, March 13, 2012 at 5:03pmThank you. From now on, I will include my work for reference. As you can see, I do know (to a degree) what I am doing. However, my practice instructions aren't always very clear on how to carry out these problems so I look for a second opinion to back me up. Again, thank you.

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