Finding a mistake in the computation of a double integral in polar coordinates I have to find $P\left(4\left(x-45\right)^2+100\left(y-20\right)^2\leq 2 \right) $
$f(x)$ and $f(y)$ are given, which I will use in my solution below .
$$P\left(4\left(x-45\right)^2+100\left(y-20\right)^2\leq 2 \right) = \int\int_D f(x)\cdot f(y)\hspace{1mm}dydx$$
Where D is the region inside $4\left(x-45\right)^2+100\left(y-20\right)^2\leq 2$
Substitute $x = 45+\dfrac{r}{2}\cos\theta$ and $y = 20+\dfrac{r}{10}\sin\theta$
The Jacobian will be $r\cdot \dfrac{1}{2}\cdot \dfrac{1}{10} = \dfrac{r}{20}$
Then we can say $D \in \left( r, \theta\right)\hspace{1mm}|\hspace{2mm} 0\leq r\leq \sqrt{2},\hspace{2mm}0\leq \theta \leq 2\pi$
$$\int\int_D f(x)\cdot f(y)\hspace{1mm}dydx = \int_0^{2\pi} \int_{0}^{\sqrt{2}}\left[ \dfrac{2}{\sqrt{2\pi}}\hspace{1.4mm} e^{-0.5r^2\cos^2\theta}\right]\cdot \left[\dfrac{10}{\sqrt{2\pi}} e^{-0.5r^2\sin^2\theta} \right] \cdot \left[ \dfrac{r}{20}\right]drd\theta$$
$$= \int_0^{2\pi} \int_{0}^{\sqrt{2}}\dfrac{r}{2\pi}\hspace{1.4mm} e^{-0.5r^2\cos^2\theta-0.5r^2\sin^2\theta}\hspace{1mm}drd\theta$$
$$= \dfrac{1}{2\pi}\left[\int_0^{2\pi} d\theta \right]\left[\int_{0}^{\sqrt{2}}r\hspace{1.4mm} e^{-0.5r^2}\hspace{1mm}dr \right]$$
Substitute $-0.5r^2 = u$ and $-r dr = du$
$$\text{Limits of Integration will change from $\int_0^{\sqrt{2}}$ to $\int_{-0^2}^{-(\sqrt{2})^2} = \int_0^{-2}$}$$
$$= \dfrac{1}{2\pi}\left[{2\pi}  \right]\left[-\int_{0}^{-2}e^{u}\hspace{1mm}du \right]$$
$$=\left[-e^{u}\right]_{0}^{-2} = 1-e^{-2} = 1-\dfrac{1}{e^2}\approx 0.8647$$
Answer at the back of the book is $0.632$!

Upper limit $-\color{red}{0.5}(\sqrt2)^2$.
