# finding the area using double integrals

• Jan 16th 2009, 09:25 AM
lllll
finding the area using double integrals
Given $r=\cos(\theta)$ and $r=\sin(\theta)$ find the area inside both circle.

Graphing it out gave me two that intersect in the first quadrant, now I was thinking that it would be

$\int_0^{\pi/2} \int_{\sin(\theta)}^{\cos(\theta)} r \ dr \ d\theta$

but this gives me $\frac{1}{8}$, whereas the solution in the back of the book tells me it's supposed to be $\frac{\pi-2}{8}$
• Jan 16th 2009, 09:38 AM
Mush
Quote:

Originally Posted by lllll
Given $r=\cos(\theta)$ and $r=\sin(\theta)$ find the area inside both circle.

Graphing it out gave me two that intersect in the first quadrant, now I was thinking that it would be

$\int_0^{\pi/2} \int_{\sin(\theta)}^{\cos(\theta)} r \ dr \ d\theta$

but this gives me $\frac{1}{8}$, whereas the solution in the back of the book tells me it's supposed to be $\frac{\pi-2}{8}$

There are two points of intersection in that quadrant. One of them is clearly 0, and the other angle at which they intersect is a solution to the equation:

$\cos{(\theta)} = \sin{(\theta)}$.

Solve that, then plug your solution into the original equations $r = \cos{(\theta)}$ or $r = \sin{(\theta)}$. This will give you the REAL r limits.
Hint...

$\cos{(\theta)} = \sin{(\theta)}$

Divide through by $\cos{(\theta)}$

$1 = \tan{(\theta)}$

(Answer should be $0 \leq r \leq \frac{\sqrt{2}}{2}$)
• Jan 16th 2009, 12:00 PM
lllll
I see that the they interest at those given points, but shouldn't my area be bounded by the given equations?
• Jan 16th 2009, 01:41 PM
Mush
Quote:

Originally Posted by lllll
Given $r=\cos(\theta)$ and $r=\sin(\theta)$ find the area inside both circle.

Graphing it out gave me two that intersect in the first quadrant, now I was thinking that it would be

$\int_0^{\pi/2} \int_{\sin(\theta)}^{\cos(\theta)} r \ dr \ d\theta$

but this gives me $\frac{1}{8}$, whereas the solution in the back of the book tells me it's supposed to be $\frac{\pi-2}{8}$

BTW, how did you manage to get $\frac{1}{8}$ anyway? By my reckoning:

$\displaystyle \int_0^{\frac{\pi}{2}} \int_{\sin(\theta)}^{\cos(\theta)} r \ dr \ d\theta$

$\displaystyle = \int_0^{\frac{\pi}{2}} [\frac{r^2}{2}]_{\sin(\theta)}^{\cos(\theta)} \ d\theta$

$\displaystyle = \int_0^{\frac{\pi}{2}} \frac{\cos^2{\theta}}{2} -\frac{\sin^2{\theta}}{2} d\theta$

$\displaystyle = \int_0^{\frac{\pi}{2}} \frac{\cos^2{\theta} - \sin^2{\theta}}{2} d\theta$

$\displaystyle = \int_0^{\frac{\pi}{2}} \frac{\cos{2\theta}}{2} d\theta$

$\displaystyle = \frac{1}{4} [\sin{2\theta}]_0^{\frac{\pi}{2}}$

$\displaystyle = \frac{1}{4} (\sin{\pi} - \sin{0}) = 0$

Clearly your integral was not correct to begin with, but neither was your integration process!
• Jan 16th 2009, 02:17 PM
lllll
I must have made a mistake in my calculation, but I'm still no were near my final answer.

I've tried :

$\int_0^{\pi/2} \int_0^{\sqrt{2}/2} tan(\theta) \cdot r \ dr \ d\theta$ = undifined

$\int_0^{\pi/2} \int_0^{\sqrt{2}/2} r \ dr \ d\theta =frac{1}{8} \pi$

$\int_0^{\pi/2} \int_{\cos(\theta)}^{\sin(\theta)} tan(\theta) \cdot r \ dr \ d\theta$ = undifended

even changing $\frac{\pi}{2} \rightarrow \pi \ \mbox{or} \ 2\pi$ got me nowhere
• Jan 16th 2009, 02:33 PM
galactus
Try this:

$2\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_{0}^{cos ({\theta})}rdrd{\theta}$

See why?. The region shaded in the graph is half the region needed, so we multiply by 2 since there is symmetry.
• Jan 16th 2009, 08:33 PM
DeMath
Quote:

Originally Posted by lllll
Given $r=\cos(\theta)$ and $r=\sin(\theta)$ find the area inside both circle.

Graphing it out gave me two that intersect in the first quadrant, now I was thinking that it would be

$\int_0^{\pi/2} \int_{\sin(\theta)}^{\cos(\theta)} r \ dr \ d\theta$

but this gives me $\frac{1}{8}$, whereas the solution in the back of the book tells me it's supposed to be $\frac{\pi-2}{8}$

This is a graphical representation of your problem in polar coordinates.