Answer to Question #229405 in Chemical Engineering for Lokika

Question

By Changing the order of Integration and evaluate ∫0^{a}∫0^{√a^{2}-y^{2}}(x^{2}+y^{2})dxdy `

Answer 1

$∫_0^{a}∫_0^{\sqrt{a^{2}-y^{2}}}(x^{2}+y^{2})dxdy \\ ∫_{y=0}^{y=a}∫_{x=0}^{x=\sqrt{a^{2}-y^{2}}}(x^{2}+y^{2})dxdy \\$

From the above case, limits of integration are

$x= 0 \space to \space x= \sqrt{a^2-y^2}\\ y= 0 \space to \space y= a\\ x= \sqrt{a^2-y^2} \implies y= \sqrt{a^2-x^2}\\$

So new limits of integration are

$∫_{x=0}^{x=a}∫_{y=0}^{y=\sqrt{a^{2}-x^{2}}}(x^{2}+y^{2})dydx \\ ∫_{0}^{a}∫_{0}^{\sqrt{a^{2}-x^{2}}}(x^{2}+y^{2})dydx \\ =\int _0^{\sqrt{a^2-x^2}}x^2dy+\int _0^{\sqrt{a^2-x^2}}y^2dy\\ =\int _0^a\left(x^2\sqrt{a^2-x^2}+\frac{\left(a^2-x^2\right)^{\frac{3}{2}}}{3}\right)dx\\ =\int _0^ax^2\sqrt{a^2-x^2}dx+\int _0^a\frac{\left(a^2-x^2\right)^{\frac{3}{2}}}{3}dx\\ =\frac{\pi }{16}a^4+\frac{\pi a^4}{16}$