1. change of variable

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

My solution:

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

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

My solution:

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

$\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 $x^3+y^3=(x+y)(x^2-xy+y^2)$

so you get

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

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

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

Good luck

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

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

is my solution still holds?

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

is my solution still holds?
yes