Question

Change variables and evaluate the new integral.

\[\int\limits_{}^{}\int\limits_{R}^{}x ^{2}\sqrt{x+2y}dA\] where \[R=\left\{ (x,y): 0 \le x \le 2, \frac{ -x }{ 2 } \le y \le 1-x\right\}\]
Use\[x=2u\] and \[y=v-u.\]

Answer 1

I've already figured out the new limits of integration which are \[0 \le u \le 1\]\[0 \le v \le 1-u\] and the Jacobian which is 2.

Answer 2

where are you stuck then?

Answer 3

I guess the integration.

Answer 4

so you can setup the integral?

Answer 5

\[\int\limits_{0}^{1}\int\limits_{0}^{1-u} 4u ^{2}\sqrt{2v} \times 2 dvdu\]

Answer 6

ok

Answer 7

\[8\sqrt{2}\int\limits_{0}^{1}\left[u^2\int\limits_{0}^{1-u} \sqrt{v} dv\right]du\]

Answer 8

two calc I integrals

Answer 9

I get to \[\frac{ 16\sqrt{2} }{ 3 }\int\limits_{0}^{1} u ^{2}(1-u)^{\frac{ 3 }{ 2 }} du\]

and I get stuck!

Answer 10

substitution \(w=1-u\)

Answer 11

Then what does u^2 become?

Answer 12

\[w=1-u\]

\[u=1-w\]

\[u^2=(1-w)^2\]

Answer 13

Well that makes sense! Thanks!

Answer 14

Ok but I don't know how to evaluate that integrand either??

Answer 15

\[(1-w)^2w^{3/2}\]

\[(1-2w+w^2)w^{3/2}\]

\[w^{3/2}-2w^{5/2}+w^{7/2}\]

Answer 16

Ok that helps!

Answer 17

So the answer is\[\frac{ 256\sqrt{2} }{ 945 }\] Thank you!

Answer 18

yep