Calculating the volume of an elliptical region

Question

Problem:

Calculate the volume of the finite body $K$ that is limited by the surfaces

$$ z=2-x^2-y^2 \\z=y^2 $$

Answer: $\pi\sqrt2$

My Attempt:

The surfaces intersect when $x^2+2y^2 = 2$

Therefore, the volume is:

$$ \iint_{A}(2-x^2-y^2)-(y^2)=\iint_{A}2-x^2-2y^2 $$

where $A : x^2+2y^2 \le 2$.

The variable substitution $$ u = x \\v = \sqrt2y $$

transforms the ellipsis into a circle, and $\frac{d(x, y)}{d(u,v)} = \frac{1}{\sqrt2}$.

Consequently, the volume can be written as:

$$ \iint_{B}(2-u^2-v^2)\frac{1}{\sqrt2}\,dudv $$

where $B : u^2+v^2 \le 2$

Polar coordinates can now be used, and we finally have the volume:

$$ \iint_{B}(2-u^2-v^2)\frac{1}{\sqrt2}\,dudv=\int_{0}^{2\pi}\int_{0}^{2}(2-r^2)\frac{r}{\sqrt2}drd\theta = 0 $$

which is clearly wrong! Where do I run off the rails?

Answer 1

The problem lies in $\displaystyle\int_0^2$. In fact, $u^2+v^2\leqslant2\iff\sqrt{u^2+v^2}\leqslant\sqrt2$, and therefore it should be $\displaystyle\int_0^{\sqrt2}$.