Volume of a Surface of Revolution

**onlinemathstudent** · June 7th 2011, 02:06 PM

I am working on a problem and wanted to confirm the result I got:

The function f(x) = (3/50)x^5 bounded by x=2 revolved about the y-axis.

I set up the integral as follows:

$\int ^{48/25} _{0} \pi {\left( {\left( \frac{50}{3 }\right)}^{\frac{1}{5 }} {y}^{\frac{1}{5 } }\right)} ^{2 } dy$

My answer is:

$\frac{192}{ 35}\pi$

Did I get it right?

**Ackbeet** · June 7th 2011, 06:03 PM

Originally Posted by onlinemathstudent

I am working on a problem and wanted to confirm the result I got:

The function f(x) = (3/50)x^5 bounded by x=2 revolved about the y-axis.

I set up the integral as follows:

$\int ^{48/25} _{0} \pi {\left( {\left( \frac{50}{3 }\right)}^{\frac{1}{5 }} {y}^{\frac{1}{5 } }\right)} ^{2 } dy$

My answer is:

$\frac{192}{ 35}\pi$

Did I get it right?

Apparently, you did.

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