Question #229405

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

Expert's answer

"\u222b_0^{a}\u222b_0^{\\sqrt{a^{2}-y^{2}}}(x^{2}+y^{2})dxdy \\\\\n\u222b_{y=0}^{y=a}\u222b_{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}\\\\\ny= 0 \\space to \\space y= a\\\\\n x= \\sqrt{a^2-y^2} \\implies y= \\sqrt{a^2-x^2}\\\\"

So new limits of integration are

"\u222b_{x=0}^{x=a}\u222b_{y=0}^{y=\\sqrt{a^{2}-x^{2}}}(x^{2}+y^{2})dydx \\\\\n\u222b_{0}^{a}\u222b_{0}^{\\sqrt{a^{2}-x^{2}}}(x^{2}+y^{2})dydx \\\\\n=\\int _0^{\\sqrt{a^2-x^2}}x^2dy+\\int _0^{\\sqrt{a^2-x^2}}y^2dy\\\\\n=\\int _0^a\\left(x^2\\sqrt{a^2-x^2}+\\frac{\\left(a^2-x^2\\right)^{\\frac{3}{2}}}{3}\\right)dx\\\\\n=\\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\\\\\n=\\frac{\\pi }{16}a^4+\\frac{\\pi a^4}{16}"

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