Graphical representation of double integral

Question

Consider the following double integral

$$\int_0^1 \int_{2y}^2 \cos(x^2) dx\ dy$$

I would have no problem sketching the region if instead of $\cos(x^2)$ I had nothing. However, with the introduction of the cosine term I am not sure the region the integral is defining. Could someone explain to me the difference that is introduced when I add the cosine term (a sketch would also be helpful, if it isn't very troublesome to draw).

Regarding the same integral, how do I invert the order of integration? Once again the $2y$ term confuses me.

Answer 1

In graphic terms, cosine plays no role (since it lays on the z axis, and we draw the region in the xy axis). About the change in the order of integration, note that instead of covering your region with $y\in [0,1]$ and $x\in [2y,0]$, equivalently, you may cover your region with $x\in [0,2]$ and $y\in [0,x/2]$, so $$\int_{0}^{1}\bigg(\int_{2y}^{2}f(x,y)dx\bigg)dy=\int_{0}^{2}\bigg(\int_{0}^{x/2}f(x,y)dy\bigg)dx.$$

Answer 2

the region.... $\int_0^1\int_{2y}^2 f(x,y) dx dy$

Start on the inside. $x$ from $2y$ to $2$. These are curves. $x = 2y, x=2$. and the outside limits, $y = 0$ and $y = 1$

the region is triangular, and the line $y=1$ is irrelevant, but it is the highest value y can take on based on the other three lines.

Flip the order of integration. What are the functions in terms of $y.$ $y = 0, y = (1/2) x , x = 0, x = 2$ What do your new limits of integration look like?