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Integrate the function $f(x,y) = xy(x^2+y^2)$ over the domain $R : { -3 <= x^2-y^2<3, 1 <=xy <=4 }$
asked Jan 5 by big boss (190 points)  
edited Jan 5 by randomisation

1 Answer

+1 vote

Let $x^2-y^2 = u$ and $xy = v$

$x^2+y^2 = \sqrt{(x^2-y^2)^2 + 4 x^2 y^2} = \sqrt{u^2 + 4 v^2}$

$f(u,v) =  v\sqrt{u^2 + 4 v^2}$

$I = \int \int f(x,y) dx dy = \int \int f(u,v)|J| du dv$

Now, $|J|^{-1} = \begin{vmatrix}
\frac{du}{dx} & \frac{du}{dy} \\
\frac{dv}{dx} & \frac{dv}{dy}  
\end{vmatrix} \\ = \begin{vmatrix}
2x & -2y\\
y & x
\end{vmatrix} = 2x^2 + 2y^2 = 2 \sqrt{u^2 + 4 v^2}$

Thus, $I = \int \int v\sqrt{u^2 + 4 v^2}\ \frac{1}{2\sqrt{u^2 + 4 v^2}} du dv \\

= \frac{1}{2}\int \int v du dv\\

= \frac{1}{2}\int_{v=1}^{4} \int_{u=-3}^{3} v dudv\\
= \frac{1}{2}\int_{v=1}^{4} vu \bigg\rvert_{u= -3}^{3}dv \\
= \frac{1}{2}\int_{v=1}^{4} 6v dv \\
= \frac{6}{4} v^2\bigg\rvert_{1}^{4} \\
= \frac{6}{4}(4^2-1^2)= \frac{45}{2}$

answered Jan 5 by randomisation (1,920 points)  
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