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$$ \int_{0}^{a} \int_{0}^{\sqrt{a^2-y^2}} xy(x^2+y^2)^\frac{3}{2}dxdy $$ any way to evaluate the integrand tried this with multiple substitution unable to evaluate

user895466
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2 Answers2

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First, the order of $dy \, dx$ is wrong.

Now regarding the solution, draw a picture of the region you are integrating. It is the first quadrant of the circle $\{(x,y) : x^2+y^2=a^2\}$. So, in polar coordinates $x=r \cos t$ , $y=r \sin t$ , $dx dy = r dr dt$ your integral is $$\int_0^a \int_0^{\pi/2} r \cos (t) r \sin (t) r^3 r dt dr = \left( \int_0^ar^6dr\right)\left(\int_0^{\pi/2}\sin(t)\cos(t) dt \right)=\frac{a^7}{7}\cdot \frac{1}{2} = \frac{a^7}{14}$$

Aitor Iribar Lopez
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  • holly dude i wasted a lot of time this thank you. I have not covered polar coordinates yet may be that's the issue is there any other way to solve without polar coordinates?? – user895466 Mar 04 '21 at 15:28
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    Probably you cold just try to integrate this by double integration using very big formulas, because i think primitives for $x(x^2+y^2)^{3/2}$ can be found, but these kind of problems are thought to be done with polar coordinates. In fact, every time you see $(x^2+y^2)^\alpha$ in an integral, you should try with polars – Aitor Iribar Lopez Mar 04 '21 at 15:38
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Without using polar coordinates,

$\displaystyle \int_{0}^{a} \int_{0}^{\sqrt{a^2-y^2}} xy(x^2+y^2)^{3/2} \ dx \ dy$

For first integral, substitute $t = x^2+y^2 \implies \frac{1}{2}dt = x dx$

Integrate $t^{3/2}$ wrt $dt$ and substitute $t$ back with $x^2+y^2$ and you get,

$\displaystyle \frac{1}{5}\int_{0}^{a} y \ \bigg[(x^2+y^2)^{5/2}\bigg]_{x=0}^{x=\sqrt{a^2-y^2}} \ dy = \frac{1}{5}\int_{0}^{a} y (a^5 - y^5) \ dy = \frac{a^7}{14}$

Math Lover
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