How do use integration to find the area under circle? I have recently come up with a question about the area under circle. I am not sure what to do as circles are not a function. I was thinking to find the area of blue + red, as shown in image, but I don't know how to integrate with circles. Can anyone provides a solution? Thank you.

 A: Not involving integration, but using triangle and circle areas directly. We can compare and recognize respective terms by integration.
By intersection of $y=x$ with the eccentric circle equation (omitted) we get
$$ x_L=y_L= \frac{3+\sqrt{ 89}}{2};\; TX= 10-7=3 ; $$
In the construction below due to symmetry one half of the required area is LTB, shown in green+ yellow parts:

$$ =\dfrac12\cdot TX\cdot y_L + \dfrac12 \cdot \dfrac{\pi}{4}\cdot XB^2 =\dfrac12\cdot TX\cdot y_L +  \dfrac{7^2}{2}\cdot\tan^{-1}\frac{y_L}{x_L-3}$$
which you can calculate numerically.
A: A sketch:
The arc $AL$ has equation $y=\sqrt{49-x^2}+3$; the arc $LB$ has equation $y=\sqrt{49-(x-3)^2}$. You can show these meet at $x=x_0:=\frac{3+\sqrt{89}}{2}$. Integrate the first function from $0$ to $x_0$, then the second from $x_0$ to $10$, using$$\int\sqrt{a^2-x^2}\;dx=\tfrac12(a^2\arcsin\tfrac{x}{a}+x\sqrt{a^2-x^2})+C.$$
A: The circle on the $x$-axis is
$(x- 3)^2 + y^2= 49$. In polar coordinates, it is
$$r (\theta) = 3\cos t +\sqrt{9\cos^2 t+40}
$$
Then, the total area is
$$A= 2\int_0^{\pi/4} \frac12r^2 (\theta)\>d\theta 
=\frac32 \left(\sqrt{89}+3\right) + 49\tan^{-1}\frac {\sqrt{89}+3}{\sqrt{89}-3}
$$
A: Find the abscissa/ordinate of the point $L$ by intersection, then compute the area of the curvilinear triangle $WLA$ by the integral
$$I=\int_0^{x_L}(\sqrt{7^2-x^2}-y_L)\,dx.$$ The answer is $2I$ plus the area of the square $T_1J_2LW$.

