Is the double integral equal to the area? I have to compute the double integral $\int_0^{\sqrt{\frac{\pi}{2}}}\int_x^{\sqrt{\frac{\pi}{2}}} 2\sin(y^2)dydx$. 
I drawed the region $0\leq x\leq \sqrt{\frac{\pi}{2}}$ and $x\leq y\leq \sqrt{\frac{\pi}{2}}$. This is a triangle with height and base equal to $\sqrt{\frac{\pi}{2}}$, right? 
The double integral is equal to the are of the triangle, or not? 
So, is the double integral equal to $\frac{1}{2}\cdot \sqrt{\frac{\pi}{2}}\cdot \sqrt{\frac{\pi}{2}}=\frac{\pi}{4}$ ?
 A: You have a heap of sand on a triangular lot. The height of the heap at the point $(x,y)$ is defined to be $2\sin(y^2)$.  The integral you are told to compute is the volume of that heap.
A: Hint. Note that the double integral over that triangle can be written as iterated integrals in two ways. 
\begin{align*}
\int_{x=0}^{\sqrt{\frac{\pi}{2}}}\left(\int_{y=x}^{\sqrt{\frac{\pi}{2}}} 2\sin(y^2)dy\right)dx
&=\int_{y=0}^{\sqrt{\frac{\pi}{2}}}\left(\int_{x=0}^{y} 2\sin(y^2)dx\right)dy\\
&=\int_{y=0}^{\sqrt{\frac{\pi}{2}}}2\sin(y^2)\left(\int_{x=0}^{y} dx\right)dy\\
&=\int_{y=0}^{\sqrt{\frac{\pi}{2}}}2y\sin(y^2)dy.
\end{align*}
Can you take it from here? 
P.S. The final result is different from what you have found. It is not the area of the triangle. 
A: The triangular region you found is called the domain of integration. The integrand function is a two variable function: $f(x,y)=2\sin(y^2)$. The double integral of a two-variable function shows the volume under the graph (surface) of the two-variable function in the given domain of integration. (Similarly, the integral of a one-variable function shows the area under the graph of the one-variable function in the given domain (interval) of integration). One of the methods of calculation of a double integral is to change the order of integration. (See here for examples). Hence you should apply this method for your double integral.
