Area of polar

**shilz222** · September 11th 2007, 04:53 PM

Find area of region that lies inside $r = 2+ \sin \theta$ and outside $r = 3 \sin \theta$ .

When I set $2+ \sin \theta = 3 \sin \theta$ I get $\theta = \frac{\pi}{2}$ .

How do I get the other bound?

**Jhevon** · September 11th 2007, 04:59 PM

Originally Posted by shilz222

Find area of region that lies inside $r = 2+ \sin \theta$ and outside $r = 3 \sin \theta$ .

When I set $2+ \sin \theta = 3 \sin \theta$ I get $\theta = \frac{\pi}{2}$ .

How do I get the other bound?

they only intersect at that one point. the second curve is completely contained in the first, so just find the area under the first curve and subtract the area under the second, that is for $0 \le \theta \le 2 \pi$

**shilz222** · September 11th 2007, 06:24 PM

I got $-1$ . is this correct?

**Jhevon** · September 11th 2007, 06:53 PM

Originally Posted by shilz222

I got $-1$ . is this correct?

no, it's a negative answer, how could it be the area? remember the formula to calculate areas in polar coordinates. $A = \frac {1}{2} \int_{\alpha}^{\beta}r^2~d \theta$

the area is given by: $\frac {1}{2} \int_{0}^{2 \pi} (2 + \sin \theta )^2 ~d \theta - \frac {1}{2} \int_{0}^{\pi} (3 \sin \theta )^2~d \theta$

**shilz222** · September 11th 2007, 06:55 PM

thats what I did, and I got -1.

**Jhevon** · September 11th 2007, 06:57 PM

Originally Posted by shilz222

thats what I did, and I got -1.

well you made a mistake somewhere, you should get $\frac {9 \pi}{4}$ . show your solution let me see where you went wrong

also note that the second integral is that of a circle with center (0, 3/2) and radius 3/2. you could find the area the old fashioned way of $A = \pi r^2$ , where r is the radius, not the polar function

**shilz222** · September 11th 2007, 06:58 PM

wait why is the second one from 0 to pi?

**Jhevon** · September 11th 2007, 07:00 PM

Originally Posted by shilz222

wait why is the second one from 0 to pi?

did you graph the curves? you would have noticed that the second curve retraces itself from $\pi$ to $2 \pi$ , so finding the limits for $0 \le \theta \le 2 \pi$ will find the area twice. you still shouldn't have ended up with -1 though. were that the case, you would have ended up with 0

**shilz222** · September 11th 2007, 07:01 PM

oh ok. Thats what i did wrong

thanks

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