# Polar coordinates area

• May 3rd 2009, 09:54 AM
kyva1929
Polar coordinates area
r=cos(t) r=sin(t)

find the area of intersection.

I come up with integrals of cos^2 t - sin^2 t from 0 to pi/4, and from 5pi/4 to 2pi, sin^2 t - cos^2 t from pi/4 to 5pi/4.

is that correct and, is there a simpler way to tackle the problem?

Thank you very much!
• May 3rd 2009, 10:33 AM
curvature
Quote:

Originally Posted by kyva1929
r=cos(t) r=sin(t)

find the area of intersection.

I come up with integrals of cos^2 t - sin^2 t from 0 to pi/4, and from 5pi/4 to 2pi, sin^2 t - cos^2 t from pi/4 to 5pi/4.

is that correct and, is there a simpler way to tackle the problem?

Thank you very much!

From the graph below, we should have

Area= $2*\frac{1}{2}\int_{0}^{\frac{\pi}{4}}sin^{2}(\thet a)d\theta$
• May 3rd 2009, 10:37 AM
kyva1929
Quote:

Originally Posted by curvature
From the graph below, we should have

Area= $2*\frac{1}{2}\int_{0}^{\frac{\pi}{4}}sin^{2}(\thet a)d\theta$

Oh! I just figure it out by plotting the graph with maple.

What do you use to plot the graph attached? It seems nicer than the one I plotted with maple.

Thank you!
• May 3rd 2009, 10:48 AM
skeeter
Quote:

Originally Posted by kyva1929
r=cos(t) r=sin(t)

find the area of intersection.

I come up with integrals of cos^2 t - sin^2 t from 0 to pi/4, and from 5pi/4 to 2pi, sin^2 t - cos^2 t from pi/4 to 5pi/4.

is that correct and, is there a simpler way to tackle the problem?

Thank you very much!

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos^2{t} \, dt$

or

$\int_0^{\frac{\pi}{4}} \sin^2{t} \, dt$
• May 4th 2009, 08:09 AM
curvature
Quote:

Originally Posted by kyva1929
Oh! I just figure it out by plotting the graph with maple.

What do you use to plot the graph attached? It seems nicer than the one I plotted with maple.

Thank you!

The graph was plotted with maple.