# Question about the definite integral $\int_0^1 \sqrt{\frac{4}{5}-x^2}{\rm d}x$

I was doing an exercise and during this exercise I had to solve the definite integral in order to calculate an area. The area was of $A=\{(x,z):x,z \geq 0, x+\frac{1}{2}z \leq 1, x^{2}+z^{2} \geq \frac{4}{5}\}$ The area is a triangle cutted by 1/4 of a circle,so the area i easy to evaluate and is $1-\frac{\pi}{5}$. I tried doing $\int_0^1dx\int_\sqrt{\frac{4}{5}-x^2}^{-2(x-1)}dz$

But I had problem with this piece:

$$-\int_0^1 \sqrt{\frac{4}{5}-x^2}{\rm d}x$$

I solved the indefinite integral by putting $x=\frac{2}{\sqrt{5}}\sin(t)$.

But now the problem is that the integration's extrems are in $\mathbb C$ and no more in $\mathbb R$.

I noticed that (but i don't no why) solving now the integral in $\mathbb C$ and than taking the real part of the result make me solve the integral. My question is: there is a way to solve this integral in $\mathbb R$? And why the way I adopted to solve the integral worked?

• what do you mean by 'it worked'? did you end up having to compute arcsin for a value greater than 1? I guess you mean you plugged complex number limits in though – Cato May 19 '17 at 10:56
• Can an area problem really give you something involving the square root of negative values? When $x$ is close to $1$, you have $4/5 - x^2 < 0$. – GEdgar May 19 '17 at 10:59
• @Cato yes, and then I took only the real part of the result – Tarlo_x May 19 '17 at 12:04
• @GEdgar I said it was only a piece of the integral, taking the real part of the piece above results -1/5pi. But the other piece gieve me 1. So the result was positive – Tarlo_x May 19 '17 at 12:07
• The integral you have put up does not represent the area you want to compute. For example the area of the upper part in the $(x,z)$ plane can be written $\int_0^{x_{\rm max}}{\rm d}x \int_{\sqrt{4/5 - x^2}}^{2-2x}{\rm d}y$ where $x_{\rm max} = 4/5$ is the point where the line and the circle intersect. Likewise the area of the lower right part is $\int_0^{y_{\rm max}}{\rm d}y \int_{\sqrt{4/5-y^2}}^{1-y/2}{\rm d}x$ where $y_{\rm max} = 2/5$. Draw a sketch to see why. – Winther May 19 '17 at 14:35

If you use the substitution $x = \frac{2}{\sqrt{5}}\sin t$, then $$t = \sin^{-1}\left(\frac{\sqrt{5}}{2}x\right).$$ Now, for $x = 1$ we have $t = \sin^{-1}\left(\frac{\sqrt{5}}{2}\right)$ which is not defined on $\mathbb{R}$