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By replacing $x=r\cos(a)$ and $y = r\sin(a)$ we get $f(x,y)$ as $r\cos^2 a\sin a$, but $r$ can be any no. Belongs to $\mathbb R$ then how come function is bounded. As we can choose $r$ any value and hence it should be unbounded.

Please correct me with explanation.

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2 Answers 2

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It's not bounded.

Because for $x\neq0$ we have $$f(x,x)=\frac{x}{2}$$

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Your explanation is also correct. $$f(x,y) =\frac{x^2y}{x^2 + y^2}$$ Now switch to polar coordinates. put $x= r cos\theta$ and $y = rsin\theta$ then $f(rcos\theta , rsin\theta) = r cos^2 \theta sin\theta$. Now for a fix values value of $\theta$ (except where $\sin\theta$ and $cos\theta$ become zero) if u will vary $r$ then your will become unbounded.

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