2
$\begingroup$

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 $0.632$!

$\endgroup$
  • $\begingroup$ Is this a probability computation? Is $f$ a density function? We can't tell if you've set the problem up correctly without more information. Can you state the problem exactly? $\endgroup$ – MPW Jul 27 '14 at 20:22
  • $\begingroup$ Yves Daoust you should get medal $\endgroup$ – Holy cow Jul 27 '14 at 20:30
  • $\begingroup$ You have asked more than 50 questions by now. Please make an effort to give them titles that indicate what the problem is about. "Where did I go wrong?" gives no clue to what the question is about. $\endgroup$ – user147263 Jul 27 '14 at 20:47
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.