Double integral with substitution polar

Question

Im faced with the followng

$$\int \int x^2y^2\,\mathrm dx\,\mathrm dy$$

D={(x,y):x^2+y^2≤1

Now the strategi I am trying to learn is using substitution for double integrals. The way when you use polar coordinates.

I think that we should substitute dxdy to rdrda

Where r is the radius of a circle and a is the angle. Then the limits are 0≤a≤2pi and 0≤r≤1.

Im not sure how to substitute x^2y^2. Can we make r = x^2y^2 or does it have to be r= x^2+y^2, since that is our boundary?

Can someone calculate this integral and explain the steps?

Thank you. Please tell if something was unclear so i can improve my questions.

/John

Answer 1

if $(x,y)=(r\cos\theta ,r\sin\theta )$, then $$x^2y^2=r^4\cos^2\theta \sin^2\theta .$$

Therefore,

$$\iint\limits_{\{(x,y)\in\mathbb R^2\mid x^2+y^2\leq 1\}}x^2y^2dxdy=\int_0^{2\pi}\int_0^1 r^5\cos^2(\theta )\sin^2(\theta )drd\theta =\int_0^1 r^5dr\int_0^{2\pi}\sin^2(\theta )\cos^2(\theta )d\theta . $$

Edit

Use the fact that $\sin(2x)=2\cos(x)\sin(x)$ and $\sin^2(x)=\frac{1-\cos(2x)}{2}$ allow you to compute $\int_0^{2\pi}\cos^2(\theta )\sin^2(\theta )d\theta .$

Answer 2

Hint:

in polar coordinates (with your notation) we have: $$ x=r\cos a \qquad y=r\sin a $$

Answer 3

Recall that in polar coordinates

• $x=r\cos \theta$
• $y=r\sin \theta$

thus

$$\int \int_D x^2y^2\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}d\theta \int_0^1 r^5\cos^2 \theta \sin^2 \theta dr=\frac1{24}\int_0^{2\pi} \sin^2 (2 \theta)\, d\theta=\\=\frac1{48}\int_0^{4\pi} \sin^2 t\, dt=\frac{\pi}{24}$$

indeed

$$\int_0^{4\pi} \sin^2 t\, dt=\int_0^{4\pi} 1-\cos^2 t\, dt\implies2\int_0^{4\pi} \sin^2 t\, dt=4\pi\implies \int_0^{4\pi} \sin^2 t\, dt=2\pi$$