Question

change to polar and then integrate

Answer 1

\[\int\limits_{0}^{2}\int\limits_{0}^{\sqrt{4-y ^{2}}} x ^{2}+y^{2} dxdy\]

Answer 2

i know it becomes \[r^{2} r dr dTheta\]

Answer 3

yes, which simplifies to\[r^3drd\theta\]where are you stuck?

Answer 4

the sqrt term becomes x^2+y^2= 4 right

Answer 5

x^2+y^2=r^2
what is the radius of the circle in question?

Answer 6

so i should get r^4/4 evaluated from 0 to 2
then i should get 4 dtheta

Answer 7

radius should be 2

Answer 8

yes

Answer 9

and what are the bounds on theta?

Answer 10

0 to 2pi

Answer 11

remember that sqrt(4-y^2) is only the *top* of the circle, the bottom would require a negative sign, so the bounds on theta are only...?

Answer 12

0 to pi

Answer 13

yes

Answer 14

so would i ge 4 pi

Answer 15

yes

Answer 16

ur a life saver

Answer 17

no problem, happy to help