Finding volume using the disc method

**yeloc** · April 7th 2009, 05:50 PM

$y=6-2x-x^{2}$
$y=x+6$

Rotate around the x-axis.

**Chris L T521** · April 7th 2009, 06:18 PM

Originally Posted by yeloc

$y=6-2x-x^{2}$
$y=x+6$

Rotate around the x-axis.

First find the points of intersection.

They intersect when $x+6=6-2x-x^2\implies x^2+3x=0\implies x(x+3)=0\implies x=0,-3$ .

Thus, by the washers, you integral will take on the form of $\pi\int_a^b\left(\left[f\!\left(x\right)\right]^2-\left[g\!\left(x\right)\right]^2\right)\,dx$

In this case $f\!\left(x\right)=6-2x-x^2$ and $g\!\left(x\right)=x+6$

Thus, the volume of the solid about the x axis is $\pi\int_{-3}^0\left(6-2x-x^2\right)^2-\left(x+6\right)^2\,dx=\pi\int_{-3}^0 x^4+4x^3-9x^2-36x\,dx$

Can you take it from here?

**yeloc** · April 7th 2009, 07:38 PM

Yes! Thanks!

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