Question

Find the areas of the region shared by the circles of the polar equations:

Answer 1

\[r = 2 \cos \theta\]
\[r = 2 \sin \theta\]

Answer 2

\[\theta=?\]

Answer 3

What do you mean? The upper and lower bounds?

Answer 4

I found them to be 0 and pi if that's what you mean.

Answer 5

yp, are they given ?

Answer 6

Nope.

Answer 7

I just used the calculator.

Answer 8

It's the same for all the questions in my book.

Answer 9

alryt which is the inner curve and the outer curve or gr8r and smaller curve?

Answer 10

2 sin theta is tangent to the x-axis and isis positive with the y-axis separating it in 2 while 2 cos theta is chopped in half by the x-axis to the left of sin theta. It is tangent to the origin.

Answer 11

r2 and r1 is ?

Answer 12

The sin theta is on top.

Answer 13

The equations are on top.

Answer 14

But the questions is to find he area shared by the two. I tried to find the area of one circle and then the area without the area of the shared. I planned to substract them to get the shared area but one of the integrations led toa negative number or a zero. So it didn't happen.

Answer 15

I was hoping if there were any other ways, especially with only one equation. But the other method would be great as long as I don't get negative or zero.

Answer 16

The top is the sin

Answer 17

\[A=\int\limits_{\alpha}^{\beta}(1/2(r _{2}^{2}-r _{1}^{2})d \theta\]

Answer 18

\[A=\int\limits_{0}^{\pi}1/2((2\cos \theta)^{2}-(2\sin \theta)^{2})d \theta\]
\[=\int\limits_{0}^{\pi}1/2(4\cos ^{2}\theta-4\sin ^{2}\theta)d \theta\]
=\[=\int\limits_{0}^{\pi}4/2(\cos ^{2}\theta-(1-\cos ^{2}\theta)d \theta\]
=\[=2\int\limits_{0}^{\pi}(2\cos ^{2}\theta-1)d \theta\]
\[=\int\limits_{0}^{\pi}(4\cos2\theta)d \theta\]

Answer 19

That's the problem. Integrating this makes the answer 0.

Answer 20

\[=\int\limits_{0}^{\pi}(2\cos2\theta)d \theta\]

Answer 21

It's 4 sin 2 theta. Plugging pi and 0 would make it zero.

Answer 22

sin 2 pi and sin 0 are both 0.

Answer 23

\[(\sin2\pi)-0 +c\]

Answer 24

yh it's not symmetrical ? it's could be from the intrsc pts

Answer 25

alpha=pi/6