Find shaded area of a segment in a circle

Question

A circle centered at the origin is of radius $10$ units. Point $A = (10, 0)$ and point $B = (0, 10)$. Point $C = (-5, 5)$. I want to find the area of the region with sides $CA$ and $CB$ and arc $AB$. This is the shaded area in the figure below.

My attempt:

The parametric equation of the circle is

$ P(t) = 10 (\cos t , \sin t ) $

Therefore, the vector $r(t)$ extending from $C$ to $P(t)$ is

$r(t) = (x(t), y(t)) = P(t) - C = (5, -5) + 10 (\cos t , \sin t ) $

The area is given by

$ \text{Area} = \dfrac{1}{2} \displaystyle \bigg| \int_{t = 0}^{t = \dfrac{\pi}{2}} x(t) y'(t) - x'(t) y(t) \ d t \bigg| $

The derivative of $r(t)$ is

$r'(t) = (x'(t), y'(t) ) = 10 ( - \sin t, \cos t ) $

Therefore,

$ \text{Area} = \dfrac{1}{2} \displaystyle \bigg| \int_{t=0}^{t = \dfrac{\pi}{2}} (5 + 10 \cos t)(10 \cos t ) - (-10 \sin t )(-5 + 10 \sin t ) \ d t \bigg| $

This simplifies to

$ \text{Area} = \dfrac{1}{2} \displaystyle \bigg| \int_{t=0}^{t = \dfrac{\pi}{2}} 100 + 50 \cos t + -50 \sin t \ d t \bigg| $

This evaluates to,

$ \text{Area} = \dfrac{1}{2} ( 50 \pi ) = 25 \pi $

I'd like to verify if I got the right answer.

Your comments, hints, or alternative answers are highly appreciated.

Answer 1

I'll offer an alternative geometric solution. Introduce line $AB$ to split the area. Let's name the triangle's area $S_1$, and what's left over as $S_2$.

It is easy to see that $S_2$ is the difference between the quarter circle (from $A$ to $B$) and $S_{ABO}$ ($O$ is the origin). So, $$S_2=\frac{1}{4}\pi\cdot 10^2-\frac{10\cdot 10}{2}=25\pi-50$$

Now we simply need to find the area of triangle $ABC$. It's easy to check that $AB, BC$ are perpendicular (as their slopes multiply to $-1$). Thus, we can use the distance formula and find $S_1$,

$$S_1=\frac{\sqrt{(10-0)^2+(0-10)^2}\sqrt{(0+5)^2+(10-5)^2}}{2}=\frac{10\cdot 5\cdot (\sqrt {2})^2}{2}=50$$

Hence the total colored area is $S_1+S_2=25\pi$, which agrees with your solution.

Answer 2

Let the center of the circle be $O$. Line $AB$ is parallel to line $OC$. As such, the area of triangle $ABC$ is equal to the area of triangle $ABO$. The shaded area is exactly one quarter of the area of the circle. $\frac{\pi r^2}{4}=25\pi$

Answer 3

Call intersection of AC, P at $(0,10/3)$.

Shaded area in Quadrant II, triangle CPB =$ (1/2)(5)(10-10/3)=50/3 $

Unshaded area in Quandrant I, triangle PAB= $(1/2)(10/3)(10)=50/3$

Total area in Quadrant I = $(1/4)\pi(10)^2=25\pi$

As mentioned elsewhere the shaded area cancels out the unshaded area so the total shaded area is a quarter of the circle.

Alternatively:

Call point $(-10,0)$, R.

Area of triangle RCA = $(1/2)(20)(5)=50$

Area of a circle above line RB: $(1/4)\pi (10)^2- (1/2)(10)^2=25\pi-50$

Area of shaded region = $(1/2)\pi(10)^2-( A_{RCA}+A_{RB})=25 \pi$