Evaluate the double integral by changing to polar coordinates for $x^2+y^2\leq4$
Solution 1
İf you think $z=\sqrt{4x^2 y^2}$ and $0\leq y$ , it shows a semisphere. $$\int_{2}^2 \int_0^{2\sqrt{4x^2}} \int_0^\sqrt{4x^2  z^2} \, dy \, dz \, dx$$
Converting to polar coordinates in double integral;
Note:Hence there is a symmetry , we can think like $z \geq 0$ instead of $y \geq0$ $$ \int_0^{2\pi} \int_0^2 \sqrt{4r^2}\,r \, dr \, d\theta $$
Solution 2
Equation of the circle is $x^2+y^2=r^2$
Since you have $ x^2+y^2=4$
That means radius of your circle is $2$
So the following integral will become
$$\iint_D \sqrt{4x^2y^2} \, dx \, dy = \int_0^\pi \, d\theta \int_0^2\:r\sqrt{4r^2}\ dr$$
$\theta$ is from $0$ to $\pi$ because you have only upper half of a circle. $\sqrt{4r^2}$ gets multiplied on $r$ because you need to take into account $dxdy = rdrd\theta$
Lauren Bathers
Updated on August 17, 2020Comments

Lauren Bathers about 3 years
Change the double integral $\iint_D \sqrt{4x^2y^2} \, dx \, dy$ where $D = \{(x,y):x^2+y^2\leq4,y\geq0\}$ by changing to polar coordinates $r, \phi$
So am I right in thinking the limits would be $0$ and $4$ for $x$ and $y$?
Converting the integral would be
\begin{align} & \int_0^4 \int_0^4 \sqrt{4x^2y^2} \, dx \, dy = \iint_D \sqrt{4r^2\cos^2\phir^2\sin^2\phi} \ r \, dx \, dy \\[10pt] = {} & \iint_D \sqrt{4r^2} \, r \, dx \, dy \end{align}
I am unsure how to change the coordinates?

corcia candy almost 8 yearsare you sure end points of first integral? it sould be from $0$ to $2$

corcia candy almost 8 yearsand if you change coordinates, then you should change $dxdy$ to polar coordinates $drd\theta$

Pierpaolo Vivo almost 8 yearsThe integral is over the upper semicircular region of radius $2$ centered at the origin. Therefore it is equal to $\int_0^2 dr\ r\sqrt{4r^2}\int_0^\pi d\phi=8 \pi/3$. The original limits of integration should be $$ \int_{2}^2 dx\int_0^{\sqrt{4x^2}}dy\ . $$
