Calculating Area Inside Overlapping Polar Curves

[iRaid]

Homework Statement

Find the area inside both the circles r=2sinθ; r=sinθ+cosθ.
Express your answer as an integral, do not evaluate.

Homework Equations

$$\int_{\alpha}^{\beta}\frac{1}{2}(r_{1}^{2}-r_{2}^{2})d\theta$$

The Attempt at a Solution

So I set 2sinθ=sinθ+cosθ and solved for theta = ∏/4 and -3∏/4 (I think they're right, not sure)
So my integral was:
$$\int_{\frac{-3\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2}[(2sin\theta)^{2}-(sin\theta+cos\theta)^{2}]d\theta$$

 

[LCKurtz]

Homework Statement

Find the area inside both the circles r=2sinθ; r=sinθ+cosθ.
Express your answer as an integral, do not evaluate.

Homework Equations

$$\int_{\alpha}^{\beta}\frac{1}{2}(r_{1}^{2}-r_{2}^{2})d\theta$$

The Attempt at a Solution

So I set 2sinθ=sinθ+cosθ and solved for theta = ∏/4 and -3∏/4 (I think they're right, not sure)
So my integral was:
$$\int_{\frac{-3\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2}[(2sin\theta)^{2}-(sin\theta+cos\theta)^{2}]d\theta$$

Have you drawn the graphs? You are going to need to do that to get this problem correct. Here's one thing you need to worry about. While both graphs go through the origin, they don't do it for the same value of $\theta$. And you have negative values for $r_1$ in that range. And you don't want to count areas in common to the two regions twice.

 

[iRaid]

Would it be from 0 to pi/4?

 

[LCKurtz]

Would it be from 0 to pi/4?

No. Have you drawn the graphs?

 

[iRaid]

No. Have you drawn the graphs?

Yes I drew them, r=2sinθ is a circle with radius 2 at (0,1) and r=sinθ+cosθ is a circle that goes from (1,0) to (1,∏/2) to (0,3∏/2) and back to (1,2∏).

 

[LCKurtz]

OK, that's a start. Note that you can get the $r = 2\sin\theta$ graph for $\theta$ from $0$ to $\pi$ with $r\ge 0$. And you can get the second complete graph, with $r$ positive for $\theta$ from $-\frac \pi 4$ to $\frac {3\pi} 4$. This matters for two reasons:
1. You want $r$ nonnegative if you are going to use the formula$$
A = \frac 1 2\int_\alpha^\beta r_{outer}^2-r_{inner}^2\, d\theta$$because you want the graphs to be in the same quadrant. If either of the $r$ values are negative, they aren't.

2. The $\theta$ ranges are different. This means that you can't use that formula directly as a single integral to set up the problem anyway.

So think about this. On your graph, what area would you be calculating if you did the integral$$
\frac 1 2 \int_{-\frac \pi 4}^{\frac \pi 4}(\sin\theta +\cos\theta)^2\, d\theta$$Once you know that, do you see how to calculate the remaining area without duplicating anything?