A Question about Conditional Problability with a Joint PDF

Question

Please consider the problem below and my attempt to solve it.
Problem:
Consider a bivariate r.v $(X,Y)$ with joint pdf \begin{eqnarray*} f_{XY}(x,y) &=& \frac{1}{2\pi \sigma^2}e^{-(x^2+y^2)/(2\sigma^2))} \,\,\,\, -\infty < x, y < \infty\\ \end{eqnarray*} Find $P[(X,Y)| x^2 + y^2 \leq a^2 ]$.
Answer:
Let $K = \frac{1}{2\sigma^2}$. \begin{eqnarray*} f_{XY}(x,y) &=& \frac{K}{\pi}e^{-K(x^2+y^2))} \,\,\,\, -\infty < x, y < \infty \\ P( x^2 + y^2 < a^2 ) &=& \int_{-a}^{a} \int_{-\sqrt{ a^2 - y^2 } }^{ \sqrt{ a^2 - y^2 } } \frac{K}{\pi}e^{-K(x^2+y^2))} \, dy \, dx \\ P( x^2 + y^2 < a^2 ) &=& \int_{-a}^{a} \frac{-1}{\pi}e^{-K(x^2+y^2)} \Big|_{y = -\sqrt{a^2 - x^2} }^{ y = \sqrt{a^2 - x^2} } \, dx \\ P( x^2 + y^2 < a^2 ) &=& \int_{-a}^{a} \frac{-1}{\pi}e^{-K(x^2 + a^2 - x^2)} + \frac{1}{\pi}e^{-K(x^2 - (a^2 - x^2 )))} \\ P( x^2 + y^2 < a^2 ) &=& \int_{-a}^{a} \frac{e^{-K(2x^2-a^2)} - e^{-Ka^2})}{\pi} \\ \end{eqnarray*}
Based upon the group's comments I have updated my post.
Now, I will try again using polar coordinates. Let $K = \frac{1}{2\sigma^2}$ \begin{eqnarray*} f_{XY}(x,y) &=& \frac{K}{\pi}e^{-K(x^2+y^2))} \,\,\,\, -\infty < x, y < \infty \\ P( x^2 + y^2 < a^2 ) &=& \int_{-a}^{a} \int_{-\sqrt{ a^2 - y^2 } }^{ \sqrt{ a^2 - y^2 } } \frac{K}{\pi}e^{-K(x^2+y^2))} \, dy \, dx \\ P( x^2 + y^2 < a^2 ) &=& \int_{-\pi}^{\pi} \int_{0 }^{ a } \frac{K}{\pi}e^{-K(r^2))} r \, dr \, d\theta \\ P( x^2 + y^2 < a^2 ) &=& \int_{-\pi}^{\pi} -\frac{1}{2\pi}e^ {-Kr^2 } \Big|_{r = 0}^{r = a} \, d\theta \\ P( x^2 + y^2 < a^2 ) &=& \int_{-\pi}^{\pi} -\frac{1}{2\pi}e^{-Ka^2} + \frac{1}{2\pi}e^{0} \, d\theta \\ P( x^2 + y^2 < a^2 ) &=& \int_{-\pi}^{\pi} \frac{1}{2\pi}(1 - e^{-Ka^2} ) \, d\theta \\ P( x^2 + y^2 < a^2 ) &=& 1 - e^{-Ka^2} \\ \end{eqnarray*} Now, I realize what the book is asking. So, we have: \begin{eqnarray*} P[(X,Y)| x^2 + y^2 \leq a^2 ] &=& P( x^2 + y^2 < a^2 ) \\ P[(X,Y)| x^2 + y^2 \leq a^2 ] &=& 1 - e^{-\frac{a^2}{2 \sigma^2}} \\ \end{eqnarray*} This is the answer in the back of the book. Therefore, I feel I have done it correctly. I am hoping somebody could confirm that my answer is correct.
Thanks,
Bob