1. ## change of variable

Double integral $\displaystyle \int \int_{D} (x^3 + xy^2) dx dy$ where D is the ring domain $\displaystyle 1 \leq x^2 +y^2 \leq 2$

My solution:

$\displaystyle x=r cos\theta$
$\displaystyle y=r sin \theta$

$\displaystyle \int_{0}^{2 \pi} \int_{1}^{2} (r^3 cos^3 \theta +r cos \theta r^2 sin^2 \theta) r dr d\theta =0$

2. Originally Posted by colorful
Double integral $\displaystyle \int \int_{D} (x^3 + y^3) dx dy$ where D is the ring domain $\displaystyle 1 \leq x^2 +y^2 \leq 2$

My solution:

$\displaystyle x=r cos\theta$
$\displaystyle y=r sin \theta$

$\displaystyle \int_{0}^{2 \pi} \int_{1}^{2} (r^3 cos^3 \theta +r cos \theta r^2 sin^2 \theta) r dr d\theta =0$
remember that $\displaystyle x^3+y^3=(x+y)(x^2-xy+y^2)$

so you get

$\displaystyle \cos^3(x)+\sin^3(x)=(\cos(x)+\sin(x))(\cos^2(x)-\cos(x)\sin(x)+\sin^2(x))=$

$\displaystyle (\cos(x)+\sin(x))(1-\sin(x)\cos(x))$

$\displaystyle \int_{}^{}\int_{}^{} r^4[\sin(\theta)+\cos(\theta)-\sin^2(\theta)\cos(\theta)-\sin(\theta)\cos^2(\theta)]d\theta dr$

Good luck

3. Originally Posted by TheEmptySet
remember that $\displaystyle x^3+y^3=(x+y)(x^2-xy+y^2)$

Good luck
OMG, I have typed the wrong integral, the question sould be $\displaystyle \int \int_{D} (x^3 + xy^2) dx dy$

is my solution still holds?

4. Originally Posted by colorful
OMG, I have typed the wrong integral, the question sould be $\displaystyle \int \int_{D} (x^3 + xy^2) dx dy$

is my solution still holds?
yes