I need Integral help please

**Darkhrse99** · July 22nd 2009, 12:16 PM

I need to setup and evaluate the integral that gives volume of the solid formed by revolving around the region about the y-axis.

$x=-y^2+4y$

Volume =Pi r^2 So the Intergral evaluated from 4 to 1(guessing)V=Pi $(-y^2+4y)^2dy$ after the algebra I get $y^4-8y^3+16y^2$ My question is do I use $y^5/5-8y^4/4+16y^3/3$ to evaluate from 1 to 4?

**Chris L T521** · July 22nd 2009, 12:21 PM

Originally Posted by Darkhrse99

I need to setup and evaluate the integral that gives volume of the solid formed by revolving around the region about the y-axis.

$x=-y^2+4y$

Volume =Pi r^2 So the Intergral evaluated from 4 to 1(guessing)V=Pi $(-y^2+4y)^2dy$ after the algebra I get $y^4-8y^3+16y^2$ My question is do I use $y^5/5-8y^4/4+16y^3/3$ to evaluate from 1 to 4?

You're right...except for one thing.

It should be $\pi\int_{{\color{red}0}}^4\left(4y-y^2\right)^2\,dy=\dots$

**Darkhrse99** · July 22nd 2009, 12:57 PM

Originally Posted by Chris L T521

You're right...except for one thing.

It should be $\pi\int_{{\color{red}0}}^4\left(4y-y^2\right)^2\,dy=\dots$

So I do use the equation y5/5 ... to evaluate from 1 to 4?

**Chris L T521** · July 22nd 2009, 01:03 PM

Originally Posted by Darkhrse99

So I do use the equation y5/5 ... to evaluate from 1 to 4?

You evaluate $\tfrac{1}{5}y^5-2y^4+\tfrac{16}{3}y^3$ from 0 to 4.

**Darkhrse99** · July 22nd 2009, 01:10 PM

I got an answer of $6272Pi/15$ is this correct?

**Chris L T521** · July 22nd 2009, 01:19 PM

Originally Posted by Darkhrse99

I got an answer of $6272Pi/15$ is this correct?

No...that's quite too big...xD

Note that $\pi\int_0^4\left(4y-y^2\right)^2\,dy=\pi\left.\left[\frac{1}{5}y^5-2y^4+\frac{16}{3}y^3\right]\right|_0^4=\pi\left[\frac{1}{5}(4)^5-2(4)^4+\frac{16}{3}(4)^3\right]=\frac{512}{15}\pi$

**Darkhrse99** · July 22nd 2009, 01:24 PM

Thanks again Chris. I had it right the first time, then I redid it and came out with that huge answer.LOL

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