Question

Change the cartesian integral into an equivalent polar integral. Then evaluate the polar integral.

Answer 1

\[\int\limits_{0}^{1}\int\limits_{0}^{\sqrt{1-Y ^{2}}}x ^{2}+y ^{2}dxdy\]

Answer 2

answer is pi/8

Answer 3

i got pi/4

Answer 4

oh show ur work then if possible :P

Answer 5

I believe the graph for sqrt(1-y^2) looks like this...

Answer 6

therefore the bounds for theta should be -pi/2 to pi/2

Answer 7

the circle has a radius of one so the bounds for r i say are 0-1

Answer 8

and then x^(2)+y^(2) is equal to r^2. so that changes to that

Answer 9

and so my double integral is...

Answer 10

\[\int\limits\limits_{-\pi/2}^{\pi/2}\int\limits\limits_{0}^{1}r ^{3}drd \theta\]

Answer 11

and it's r^3 because the formula says there is always an extra r inside the double integral when you switch from cartesian to polar

Answer 12

bounds for \(\theta\) are wrong

Answer 13

what are they supposed to be?

Answer 14

because you're misinterpreting the region of integration

Answer 15

the region is not a semicircle

Answer 16

it is that is the equation for a semi-circle