Question

Calculus 3:Evaluate \[\int \int \frac{2xy}{x^2 +y^2}dxdy\] where D is the intersection of the annulus \(1 \le x^2 +y^2 \le 2\) with the second quadrant \([x \le 0 \ , y \ge 0]\) by changing to polar cordinates

Answer 1

\[x = r\cos(\theta) \ , y = r\sin(\theta) \ , \ x^2 + y^2= r^2\]

Answer 2

Right? So all I have to do is substitute these in to my function..

Answer 3

Show me your integral.

Answer 4

\[

Answer 5

I'm still trying to figure out my limits, haha.

Answer 6

\[\int_{\pi/2}^{\pi} \int_{1}^{2} \frac{2(r\cos(\theta))(r\sin(\theta))}{r^2}r dr d\theta\]

Answer 7

Is that right?... I'm just basing it off my first instinct atm.

Answer 8

Ye,s it's right.

Answer 9

dx should be reaplced with -rsin(theta)d(theta)

Let x = rcostheta
dx/dtheta = -rsintheta
therefore: dx = -rsintheta * dtheta

So it should be: -2r^2 * cos(theta) *(sin(theta))^2

Answer 10

What about \(dr\) or \(dy\)? Your substitutions wouldn't account for them.

Answer 11

\[\int_{\pi/2}^{\pi} \int_{1}^{2} \frac{2(r\cos(\theta))(r\sin(\theta))}{r^2}r dr d\theta\]\[\int_{\pi/2}^{\pi}\int_{1}^{2} \frac{2r^2(\cos(\theta))(\sin(\theta))}{r^2}r dr d\theta\]\[2\int_{\pi/2}^{\pi}\int_{1}^{2}\cos(\theta)\sin(\theta)r dr d(\theta)\]\[2\int_{\pi/2}^{\pi} \cos(\theta)\sin(\theta)d\theta \cdot \int_{1}^{2}rdr\]

Answer 12

Then I'd just use a u-sub most likely..

Answer 13

\[\overbrace{\sin\theta}^{u}\quad\overbrace{\cos\theta\;d\theta}^{du}
\]

Answer 14

Ahh alright.

Answer 15

Right now I'm looking at:
(2xy/x^2 + y^2)dx.
I am completely ignoring dy right now because for nested partial integrations, we go in => out, so I will start with the dx integration.

I ignore r when I derive dx because this is a partial integration. When I derive r with respect to theta, I treat it as a constant, so I ignore it. When I'm doing integration using dy/dr, I will consider it in my integration while treating my theta components as a constant.

Answer 16

\[2\int_{\pi/2}^{\pi} \cos(\theta)\sin(\theta)d\theta \cdot \int_{1}^{2}rdr\]\[ u = sin\theta \ , \ du = cos(\theta)d\theta \]\[2\int_{1}^{-1}u du \cdot \left.\frac{1}{2}r^2\right]_{1}^{2}\]

Answer 17

Sometimes people like to use \[
2\sin\theta\cos\theta = \sin(2\theta) = -\frac{d}{d\theta }\frac{\cos(2\theta)}{2}
\]