Question

compute the double integral y^2 dA where D is the region in the first quadrant which is enclosed by the curve defined by the equation r^2 = sin(2θ).

Answer 1

not sure how to set up integral ... its from \[\int\limits_{o}^{\pi/2}\]

Answer 2

then not sure how to set up the limits for the second one

Answer 3

Use double angle formula on the \(\sin(2\theta)\)

Answer 4

hmmm

Answer 5

Well, actually let me think for a moment.

Answer 6

I suppose you could just keep it in polar coords and convert the integrand, right?

Answer 7

wouldnt that make it more complicated

Answer 8

being r^2= 2sinxcosx

Answer 9

\[
y^2\to r^2\sin^2(\theta)
\]\[
dA=dxdy\to rdrd\theta
\]

Answer 10

yes thats what i have but how would i go about getting the lower and upper limits for the integral of dr

Answer 11

Well, the point before was, that equation is just\[
r^2=\frac{2r\sin\theta \cdot r\cos\theta }{r^2}\to x^2+y^2=\frac{2xy}{x^2+y^2}
\]

Answer 12

The limits would just be \([0,r(\theta)]\)

Answer 13

okay you lost me lol

Answer 14

The limits of \(r\) start at \(0\) and go to \(r\) in terms of \(\theta\) typically. It depends how exactly they define the region though. You should draw the region.

Answer 15

i should explain where am lost am lost where u got r^2=2rsin{theta}rcos{theta}/r^2

Answer 16

Multiplied top and bottom by \(r^2\). The point was to show how it would be converted to \(x,y\)

Answer 17

i see the conversion

Answer 18

Regardless, try to draw it.

Answer 19

its a loop on first quadrant

Answer 20

thaks

Answer 21

Did you figure it out?