∫ 0 a ∫ 0 a 2 − y 2 ( x 2 + y 2 ) d x d y ∫ y = 0 y = a ∫ x = 0 x = a 2 − y 2 ( x 2 + y 2 ) d x d y ∫_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 \\ ∫ 0 a ∫ 0 a 2 − y 2 ( x 2 + y 2 ) d x d y ∫ y = 0 y = a ∫ x = 0 x = a 2 − y 2 ( x 2 + y 2 ) d x d y
From the above case, limits of integration are
x = 0 t o x = a 2 − y 2 y = 0 t o y = a x = a 2 − y 2 ⟹ y = a 2 − x 2 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}\\ x = 0 t o x = a 2 − y 2 y = 0 t o y = a x = a 2 − y 2 ⟹ y = a 2 − x 2
So new limits of integration are
∫ x = 0 x = a ∫ y = 0 y = a 2 − x 2 ( x 2 + y 2 ) d y d x ∫ 0 a ∫ 0 a 2 − x 2 ( x 2 + y 2 ) d y d x = ∫ 0 a 2 − x 2 x 2 d y + ∫ 0 a 2 − x 2 y 2 d y = ∫ 0 a ( x 2 a 2 − x 2 + ( a 2 − x 2 ) 3 2 3 ) d x = ∫ 0 a x 2 a 2 − x 2 d x + ∫ 0 a ( a 2 − x 2 ) 3 2 3 d x = π 16 a 4 + π a 4 16 ∫_{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} ∫ x = 0 x = a ∫ y = 0 y = a 2 − x 2 ( x 2 + y 2 ) d y d x ∫ 0 a ∫ 0 a 2 − x 2 ( x 2 + y 2 ) d y d x = ∫ 0 a 2 − x 2 x 2 d y + ∫ 0 a 2 − x 2 y 2 d y = ∫ 0 a ( x 2 a 2 − x 2 + 3 ( a 2 − x 2 ) 2 3 ) d x = ∫ 0 a x 2 a 2 − x 2 d x + ∫ 0 a 3 ( a 2 − x 2 ) 2 3 d x = 16 π a 4 + 16 π a 4
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