A function $f$ is defined by $$ f(x,y) = \begin{cases} 7e^{x^2} & y \leq x \\\\ 7e^{y^2} & y > x \end{cases} $$ In the plane. Compute the double integral $$\int\int_Df(x,y)dA $$ Where $D$ is the square $[0,9] \times [0,9]$.
Im not able to integrate the function. Im also not sure what the limits should be and what "part" of the function should be where in the integral.
On
This can also be solved using cylindrical coordinates with
$$f(r,\theta) = \begin{cases} 7e^{r^2cos^2\theta} & \theta \leq \frac{\pi}{4} \\\\ 7e^{r^2sin^2\theta} & \theta >\frac{\pi}{4} \end{cases} $$
And the integral $$I = \int_{0}^{\frac{\pi}{4}} \int_{0}^{\frac{9}{cos\theta}}7e^{r^2cos^2\theta}rdrd\theta + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{0}^{\frac{9}{sin\theta}}7e^{r^2sin^2\theta}rdrd\theta = 7(e^{81}-1)$$
You should cut $D$ into two parts $D_1=\{(x,y) \in D,\ y\leq x\}$ and $D_2=\{(x,y) \in D,\ y> x\}$. For exemple, the integral over $D_1$ is (using Fubini's theorem) $$\int\int_{D_1}f(x,y)dA = 7\int_0^9\int_{0}^xe^{x^2} dydx = 7\int_0^9xe^{x^2}dx. $$ You should now be able to compute it and do the same for $D_2$.