1. ## Area of polar

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?

2. 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 $0 \le \theta \le 2 \pi$

3. I got $-1$. is this correct?

4. 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: $\frac {1}{2} \int_{0}^{2 \pi} (2 + \sin \theta )^2 ~d \theta - \frac {1}{2} \int_{0}^{\pi} (3 \sin \theta )^2~d \theta$

5. thats what I did, and I got -1.

6. 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

7. wait why is the second one from 0 to pi?

8. 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

9. oh ok. Thats what i did wrong

thanks

,

,

,

### r=2 sin(theta) r= 3 sin (theta)

Click on a term to search for related topics.