Find the area enclosed by circles $r=3\cos\theta$ and $r=3\sin\theta$

Question

Find the area enclosed by circles $r=3\cos\theta$ and $r=3\sin\theta$

I know that the graph of $r=3\sin\theta$ intersects at $\theta=0$ and at $\theta=\pi$ why aren't these two bounds used and why only $\theta$ equaling $\pi$? Also the graph $r=3\cos\theta$ intersects the pole at $\theta=\frac{3\pi}{2},\frac{1\pi}{2}$. So why do they only use $\frac{\pi}{2}$?

Answer 1

The two circles are intersecting at $\theta=\dfrac{\pi}{4}$ and $r=0$

The area region can be divided into two parts.

Note that we are bounded by the region of the red circle $($the lower part of the half$)$ by $\theta=0\mbox{ to }\theta=\dfrac{\pi}{4}$

Similarly, the region of the purple circle $($the upper part of the half$)$ is bounded by $\theta=\dfrac{\pi}{4}$ to $\dfrac{\pi}{2}$

So, we get $$\dfrac12\int_{0}^{\dfrac{\pi}{4}}(3\sin\theta)^2d\theta+\dfrac12\int_{\frac{\pi}{4}}^{\dfrac{\pi}{2}}(3\cos\theta)^2d\theta$$

Answer 2

Answer to the question: "Why do they only use $\pi/2$... ?"

Note that the intersection region is locate in the first quadrant. This is parametrized by $\theta \in \left[0,\;\frac \pi2\right]$.

Consider now a "radar ray" of an angle $\theta$ in this interval. Then it first determines an $r$-interval in the intersection domain, and we know its limits. So the integral is $$ \begin{aligned} J &=\int_{\theta \in [0,\;\pi/2]} \int_{r\in[\ 0,\; 3\min(\cos\theta,\; \sin\theta)\ ]}r\; dr\; d\theta \\ &= \int_{\theta \in [0,\;\pi/2]} \left[\frac {r^2}2\ \right]_{r=0}^{r=3\min(\cos\theta,\;\sin\theta)}\; d\theta \\ &= 2\int_{\theta \in [0,\;\pi/4]} \frac 12\cdot 9\sin^2\theta\; d\theta \\ &\qquad\text{(use the substitution $y=\pi/2-x$ for the interval $\left[\frac\pi4,\;\frac\pi2\right]$)} \\ &= \frac 98(\pi-2) \end{aligned} $$ The result fits with the geometrical intuition, we compute the double of the following:

• the area of a quarter of a disk with radius $\frac 32$,

• and from it we remove half of the area of a square with side $\frac 32$. I.e. $$ 2\left( \ \frac 14\pi\left(\frac 32\right)^2-\frac 12\left(\frac 32\right)^2\ \right)\ . $$

Answer 3

No calculus:

Let the two intersection points of the circles be A and B do that A is (0, 0) and B is (1.5, 1.5). It's easy to see that the arc formed by the center of a circle and A and B has measure $\pi/2$. So the area of that sector is $2.25\pi/4$. The area of the triangle formed by A, B, and center of the circle is $2.25/2$. So the area of half the desired area is $2.25\pi/4 -2.25/2 $ and the desired area is $5\pi/4-2.25$