$ \int_{0}^{2}\int_{0}^{4-x^2}\frac{xe^{2y}}{4-y} \,dy\,dx $

Question

Evaluate the following double integral: $$ \int_{0}^{2}\int_{0}^{4-x^2}\frac{xe^{2y}}{4-y} \,dy\,dx $$ I am not sure how to proceed and our teacher mentioned something about changing the order of integration, but I don't really know how to do this kind of problem .

Answer 1

We have

$$\int_0^2 dx \int_0^{4-x^2} \frac{xe^{2y}}{4-y} dy$$

and changing the order of integration

$$\int_0^4 dy \int_0^{\sqrt{4-y}} \frac{xe^{2y}}{4-y} dx$$

that is

$$\int_0^4 \frac{e^{2y}}{4-y}dy \int_0^{\sqrt{4-y}}x dx=\int_0^4 \frac{e^{2y}}{4-y} \left[\frac{x^2}2\right]_0^{\sqrt{4-y}}dy=\int_0^4 \frac12e^{2y} dy=\left[\frac14e^{2y} \right]_0^4=\frac14e^8-\frac14$$

Answer 2

When reversing the order of integration for a double integral, it helps to sketch the region bounded by the four equations. In this case the equations are

1. $x=0$
2. $x=2$
3. $y=0$
4. $y=4-x^2$

In the initial order of integration, the $A$ in the diagram on the $x-axis$ is moving from $0$ to $2$ [the outer integral $dx$].

When the order is reversed, the $B$ will be moving between $0$ and $4$ on the $y$-axis to form the limits of the "reversed" order with the outer integral being $dy$. So we know this much, so far:

$$ \int_0^4\int_?^? f(x,y)\,dx\,dy $$

To find the new limits of the inside integral $dx$ we have to know the coordinates of the point $C$.

Initially, they were $C=(x,4-x^2)$. But we need the coordinates of $C$ in terms of $y$. Thus we need $C=(\sqrt{4-y},y)$.

The limits of the inside integral must reflect the fact that the $x$-values are moving between the $x$-coordinate of $B$ and the $x$-coordinate of $C$. That is, between $x=0$ and $x=\sqrt{4-y}$. Thus we have:

$$ \int_0^4\int_0^\sqrt{4-y} f(x,y)\,dx\,dy $$