# Area and Arc Length in Polar Coordinates

• Oct 5th 2012, 12:11 AM
TWN
Area and Arc Length in Polar Coordinates
I'm teaching myself calculus and I can't figure out how to do the problems below. Please explain how to do them because my book is confusing me.

Find the area of the region:

1.) Inside r = 2acos(theta) and outside r = a

Find the length of the curve over the given interval:

2.) r = 2acos(theta)

3.) r = 8(1+cos(theta))

Thank you!
• Oct 5th 2012, 12:22 AM
Prove It
Re: Area and Arc Length in Polar Coordinates
Quote:

Originally Posted by TWN
I'm teaching myself calculus and I can't figure out how to do the problems below. Please explain how to do them because my book is confusing me.

Find the area of the region:

1.) Inside r = 2acos(theta) and outside r = a

Find the length of the curve over the given interval:

2.) r = 2acos(theta)

3.) r = 8(1+cos(theta))

Thank you!

It always helps to draw a picture of the graphs, so that you can determine what the terminals of your double integral should be...
• Oct 5th 2012, 12:26 AM
TWN
Re: Area and Arc Length in Polar Coordinates
How do you determine the terminals? Can it be any two angles within which the figure exists?
• Oct 5th 2012, 12:28 AM
Prove It
Re: Area and Arc Length in Polar Coordinates
Have you drawn the diagrams yet?
• Oct 5th 2012, 12:31 AM
TWN
Re: Area and Arc Length in Polar Coordinates
Yes I have. Now what do I do?

Edit: I am going to bed. If someone could have some walkthroughs ready by morning it would be greatly appreaciated.
• Oct 5th 2012, 01:29 AM
Prove It
Re: Area and Arc Length in Polar Coordinates
Don't be so rude - you don't get given walkthroughs on this site, you get given hints so that you can do your own work!

If you convert to cartesians and use horizontal strips, you'll see that the x values are bounded between \displaystyle \displaystyle \begin{align*} a - \sqrt{1-y^2} \leq x \leq \sqrt{a^2 - y^2} \end{align*} and the y values are bounded between \displaystyle \displaystyle \begin{align*} -\frac{\sqrt{3}\,a}{2} \leq y \leq \frac{\sqrt{3}\,a}{2} \end{align*}. So the double integral is

\displaystyle \displaystyle \begin{align*} \int_{-\frac{\sqrt{3}\,a}{2}}^{\frac{\sqrt{3}\,a}{2}}{ \int _{a - \sqrt{1 - y^2}}^{\sqrt{a^2 - y^2}}{\,dx}\,dy} \end{align*}