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I=  \int\limits^a_{y=0} [ {\int\limits^y_{x=0} {\sqrt{a^2-x^2}} \, dx }}] \, dy

The upper limit of integration for x, is not clear .  So I will solve for two possibilities.  Please take what ever is relevant to you.   This is a double integral. 

If the upper limit for x is "a", then the solution is simpler.

Let\ I1= \int\limits^y_0 {\sqrt{a^2-x^2}} \, dx } \\\\x=a\ Sin\phi,\ dx=a\ Cos\phi\ d\phi.\\x=0,\ then\ \phi=0;\ \ x=y,\ \phi=Sin^{-1}(\frac{y}{a})\\\\I1= \int\limits^{Sin^{-1}\frac{y}{a}}_0 {a^2\ Cos^2\phi} \, d\phi \\\\=\frac{a^2}{2}\int\limits^{Sin^{-1}\frac{y}{a}}_0 {(1+Cos2\phi)} \, d\phi \\\\=\frac{a^2}{2}[\phi+\frac{1}{2}Sin2\phi]_0^{Sin^{-1\frac{y}{a}}}\\\\If\ \phi=Sin^{-1}\frac{y}{a},\ Sin2\phi=2\frac{y}{a^2}\sqrt{a^2-y^2}\\\\I1=\frac{a^2}{2}Sin^{-1}\frac{y}{a}+\frac{a^2*2y}{4a^2}\sqrt{a^2-y^2}

If the upperlimit in the integration is a, ie., y = a then:
          I1 = πa²/4

Now the integral is :
I= \int\limits^{a}_{y=0} {\frac{\pi a^2}{4}} \, dy =\frac{\pi a^3}{4}\\

If the upper limit for integration wrt  x is not  "a", but is "y", then the solution is more complicated.  So we have :

I1=\frac{a^2}{2}Sin^{-1}\frac{y}{a}+\frac{y}{2}\sqrt{a^2-y^2}\\\\I=\frac{a^2}{2} \int\limits^a_0 {Sin^{-1}\frac{y}{a}} \, dy +\frac{1}{2} \int\limits^a_{y=0} {y\sqrt{a^2-y^2}} \, dy \\\\Let\ y=a\ Sin\phi,\ \ dy=a\ Cos\phi\ d\phi,\\y=a,\ when\ \phi=\pi/2;\ \ \ y=0,\ when\ \phi=0\\\\Let\ I2=\int\limits^a_0 {Sin^{-1}\frac{y}{a}} \, dy \\\\= \int\limits^a_0 {\phi\ *\ a\ Cos\phi} \, d\phi =a[\phi\ Sin\phi+Cos\phi]_{0}^{\frac{\pi}{2}}\\\\I2=a(\frac{\pi}{2}-1)

Let\ I3= \int\limits^a_0 {y\sqrt{a^2-y^2}} \, dy\\\\Let\ y=aSin\phi\\\\I3= \int\limits^{\frac{\pi}{2}}_0 {a\ Sin\phi\ a\ Cos\phi\ aCos\phi} \, d\phi \\\\=a^3\int\limits^{\frac{\pi}{2}}_0 {Sin\phi\ Cos^2\phi} \, d\phi \\\\=a^3[-\frac{1}{3}Cos^3\phi]_{0}^{\pi/2}\\\\I3=\frac{a^3}{3}

Finally,\ I=\frac{a^2}{2}*I2+\frac{1}{2}*I3\\\\=\frac{\pi a^3}{4}-\frac{a^3}{2}+\frac{a^3}{6}\\\\=(3\pi-4)*\frac{a^3}{12}\\
2 5 2
i hope it is clear enough. Select best answer, if u like that. click on thank you link above, pls.
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