Volume by Slicing help

**JohnM25** · February 13th 2011, 03:09 PM

Find the volume of the solid that is generated when the region bounded by

$y = (x^2/16), y = 4, x = 0$

Revolved around

$y = -2$

Can somebody tell me where to start? I'm not sure. Do I use the formula:

$\int \pi (f(x)^2 - g(x)^2) dx$

In that case, what would be the g(x)? 0?

So far I just know I'm integrating from 0 to 1.

Thanks,

John

**skeeter** · February 13th 2011, 03:49 PM

Originally Posted by JohnM25

Find the volume of the solid that is generated when the region bounded by

$y = (x^2/16), y = 4, x = 0$

Revolved around

$y = -2$

Can somebody tell me where to start? I'm not sure. Do I use the formula:

$\int \pi (f(x)^2 - g(x)^2) dx$

In that case, what would be the g(x)? 0?

So far I just know I'm integrating from 0 to 1.

Thanks,

John

first and most important step ... make a sketch of the region and the axis of rotation.

$V = \pi \int_a^b [R(x)]^2 - [r(x)]^2 \, dx$

$V = \pi \int_{-8}^8 [4 - (-2)]^2 - \left[\dfrac{x^2}{16} - (-2)\right]^2 \, dx$

**JohnM25** · February 13th 2011, 04:30 PM

Originally Posted by skeeter

first and most important step ... make a sketch of the region and the axis of rotation.

$V = \pi \int_a^b [R(x)]^2 - [r(x)]^2 \, dx$

$V = \pi \int_{-8}^8 [4 - (-2)]^2 - \left[\dfrac{x^2}{16} - (-2)\right]^2 \, dx$

Thanks, this worked out, although I integrated from 0 to 8. Seeing a picture does infact help.

John

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