Integrate the area beneath a circumference

Question

I have got the following equation:

$$(x-4)^2 + (y - 4)^2 = 16$$

I would like to find the area beneath this circumference between $x=\frac{8}{5}$ and $x = 4$

To do so, I would have to integrate $$(x-4)^2 + (y - 4)^2 = 16$$ how could I do that if that is not even a function? I mean, is there a way to integrate an equation like that?

This is the step of an algorithm that I am taking to solve the following problem:

If you guys are willing to solve, please, I would appreciate that so I would compare my result later on to see it it matches with your solutions

Thanks in advance!

Answer 1

Note that the shaded area is the difference between a trapezoid and an angle sector.

The short base of the trapezoid is $a=4-\sqrt{16-(\frac85-4)^2} = \frac45$ and the long base is $b=4$, with its height $h=4-\frac85 = \frac{12}5$. Then, the angle is given by $\sin\theta = \frac35$. Therefore, the area is,

$$I=\frac12(a+b)h - \frac12\theta b^2 = \frac{144}{25}-8\sin^{-1}\frac35$$

The geometric calculation above may be easier than yet equivalent to the area integration below

$$I =\int _{8/5}^4 y(x)dx=\int _{8/5}^4 \left( 4-\sqrt {16-(x-4)^2} \right)dx = \frac{144}{25}-8\sin^{-1}\frac35$$

Answer 2

$$ (x-4)^2+(y-4)^2=16$$

Solve for $y$

$$y=4\pm\sqrt{16-(x-4)^2}$$

The area under the lower semicircle and bounded by the lines $x=8/5$ and $x=4$ is $$A =\int _{8/5}^4 4-\sqrt {16-(x-4)^2}dx$$

You can evaluate the integral to find the answer.