[SOLVED] Double integral find volume

**atompunk** · May 10th 2010, 03:47 PM

integral from 0 to 2(integral from 0 to 1 (2x+y)^8 dx dy

My impulse is to use substitution but I am not sure if I can or not

**tonio** · May 10th 2010, 03:55 PM

Originally Posted by atompunk

integral from 0 to 2(integral from 0 to 1 (2x+y)^8 dx dy

My impulse is to use substitution but I am not sure if I can or not

$\int\limits^2_0\int\limits^1_0(2x+y)^8dxdy$ $=\frac{1}{2}\int\limits^2_0\left[\frac{(2x+y)^9}{9}\right]^1_0dy$ $=\frac{1}{18}\int\limits^2_0\left[(2+y)^9-y^9\right]dy$ $=\frac{1}{18}\left[\frac{(2+y)^{10}}{10}-\frac{y^{10}}{10}\right]^2_0=\ldots$ Why substitution if directly is pretty simple?

Tonio

**atompunk** · May 10th 2010, 05:21 PM

I think I see now. I just wasn't in the y is a constant mindset

261632/45 book answer awesome thanks. that problem kind of sucked

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