1. ## polar coordinates

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant between the circles and

i tried setting it up like this but could get the right answers

r^2 = 400
r = 20

r^2 = 20rcosΘr = 20cosΘ
i dont know how to use math tags

but my integral for dΘ would be from 0 to 2pi
and my integral for dr would be from 20cosΘ to 20i integrated the f(x,y) = 1, because that is volume, right?

2. Originally Posted by chrisc
Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant between the circles and

i tried setting it up like this but could get the right answers

r^2 = 400
r = 20

r^2 = 20rcosΘr = 20cosΘ Mr F says: This is just plain sloppy. r^2 does NOT equal 20cosΘ. r = 20cosΘ.

i dont know how to use math tags

but my integral for dΘ would be from 0 to 2pi Mr F says: What makes you think that? Have you drawn a diagram that shows the required area?

and my integral for dr would be from 20cosΘ to 20i integrated the f(x,y) = 1, because that is volume, right? Mr F says: Where is volume mentioned in the question? Nowhere that I can see. So why are you integrating wrt r as well? The question asks for an area ......Do you know the formula for the area between two polar curves?
Here is some help with answering the question you've actually posted (as opposed to the question that you might have intended to post):

The area of the region defined by $0 < r_1(\theta) < r_2(\theta)$ and $\alpha < \theta < \beta$ is given by: $\frac{1}{2} \int_{\alpha}^{\beta} \left(r_2^2 - r_1^2 \right) \, d\theta$.

For the question you have posted, $r_2 = 20$ and $r_1 = 20 \cos \theta$. And a simple sketch graph makes it very clear that $0 < \theta < \frac{\pi}{2}$.

3. r^2 = 20rcosΘr = 20cosΘ Mr F says: This is just plain sloppy. r^2 does NOT equal 20cosΘ. r = 20cosΘ.
as you can see, i wrote that. however, i accidently did not his my space bar after the theta from the first r^2 equation

and my integral for dr would be from 20cosΘ to 20i integrated the f(x,y) = 1, because that is volume, right? Mr F says: Where is volume mentioned in the question? Nowhere that I can see. So why are you integrating wrt r as well? The quesiton asks for an area ......Do you know the formula for the area between two polar curves?
there was no listed volume, thats why i said it would be 1(rdrdΘ)

that being said, i corrected my 2pi mistake an hour ago or so and it gave me the correct answer of 50pi