Creative way to find this area

Question

Let's say We have a circle with center at $(0,0)$ with radius $r$ and we have the line $y=a$ where $0 \leq a \leq r$.

the question is what is the area that between the circle and the line $y=a$(the area that above the line).

illustration for $r=5$ and $a=4$: http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2%3D25+and+y%3D4

We will end with the area as a function of $a$. So far I have tried to do integration to get the area ... it is not pretty at all.

Answer 1

Hint: The area is the difference between the areas of a sector of the circle and a triangle.

Answer 2

Consider the right triangle $(0,0), (0,a), (r \cos(\arcsin(\frac ar)),a)$ and the sector $(r \cos(\arcsin \frac ar), a), (0,r)$. Their difference in area is half the area you describe.

The area of the triangle is $\frac12 ar \cos(\arcsin \frac ar)$ and the area of the sector is $\frac{\frac\pi2-\arcsin \frac ar}2 r^2$

Answer 3

Draw a radius from the origin to the point on the circle where the line $y=a$ intersects it. Now you have the area of a sector plus the area of a right-angled triangle. Easy.

Answer 4

Using, geometry we have the equation of the circle centered at the origin as: $$x^2+y^2=r^2$$ Now, substitute $y=a$ in the above equation $$x^2+a^2=r^2 \implies x=\pm \sqrt{r^2-a^2}$$ Now, join the points of intersection $(\sqrt{r^2-a^2}, a),\,(-\sqrt{r^2-a^2}, a)$ to the origin, to get a sector with radius $r$ and an isosceles triangle with sides $r,\,r,\,2\sqrt{r^2-a^2}$

The aperture angle $\alpha$ of the sector is calculated as $$\alpha=2\cos^{-1}\left(\frac{a}{r}\right)$$ Now, the area between the circle and the line is $$ \begin{align} &\frac{1}{2}\cdot\alpha r^2-\frac{1}{2}(a)(2\sqrt{r^2-a^2})\\ =\:&\frac{1}{2}\left(2\cos^{-1}\left(\frac{a}{r}\right)\right)r^2-\frac{1}{2}a\left(2\sqrt{r^2-a^2}\right)\\ =\:&r^2\cos^{-1}\left(\frac{a}{r}\right)-a\sqrt{r^2-a^2} \end{align} $$

Where, $0\leq a\leq r$

Hence, as you mentioned for $r=5,\,a=4$. substituting these values in the expression, we get the area: $$ \begin{align} &5^2\cos^{-1}\left(\frac{4}{5}\right)-4\sqrt{5^2-4^2}\\ =\:&25\cos^{-1}\left(\frac{4}{5}\right)-12\\ \approx\:&4.08752772 \space \text{unit}^2 \end{align} $$

Answer 5

I couldn't understand what arcsin is so I have written this,
Take the angle subtended by the line at the centre of the circle. Can be done by dividing triangles into 2 halves which will be right triangles. So, $\tan\theta=\frac{upper side}{a}$ Then we can take area of the sector

Which is $\frac1{2\pi}\theta \pi r^2$

Then take the area of the triangle. $2.(A_{sector}-A_{triangle})$ will give the answer.