Change the order of integration of the following and evaluate the integral:
$$\int_{-1}^{0} \int_{-1}^y y\sqrt{x^2 + y^2} \, dx \, dy$$
I know I have to draw out the graph but im having a hard time with that… any pointers?
[Math] Changing order of integration
integrationmultivariable-calculus
Related Solutions
For (b): $\int_0^{2\pi}\int_0^1\int_z^\sqrt{2-z}rdrdzd\theta$
For (c): $\int_0^1\int_r^{2-r^2}\int_0^{2\pi}d\theta dzrdr$
You can bring $d\theta$ anywhere with other limits remaining same due to symmetry of the problem.
Please see the region shaded in the diagram.
So with change of order, the integral will be
$\displaystyle \int_1^2 \int_y^{y^2} e^{\frac{x}{y}} \ dx \ dy$
This will require you to integrate $\displaystyle y \ e^y$ in the second integral, which can be done by Integration by parts.
Edit: If you combine your second and third integral, and then with the first, you get the same integral I wrote above. You just considered change of order for each sub-region separately and so ended up with three integrals.
Best Answer
When changing the order of integration, it is very convenient to implement the integration boundary via an Iverson bracket (a method promoted by Knuth for sums), so $$\begin{align*}\int_{-1}^{0} \!dy \int_{-1}^{y} \!dx\, f(x,y) &= \int dy\int_{-1}^y\! dx\,[-1\leq y \leq 0]\,f(x,y)\\ &= \iint \!dx\,dy\, [-1\leq x\leq y\leq0]\,f(x,y)\end{align*}$$
In the second step, one can then go back and implement the integration boundaries again without the bracket. This time, however, the integration boundaries of $y$ are used first $$\begin{align*}\iint \!dx\,dy\, [-1\leq x\leq y\leq0]f(x,y) &= \int dx \int_x^{0} \!dy\, [-1\leq x \leq 0]\,f(x,y)\\ &= \int_{-1}^0\!dx\int_x^{0}\!dy\,f(x,y)\end{align*}$$
In conclusion, we have $$\int_{-1}^{0} \!dy \int_{-1}^{y} \!dx\, f(x,y) = \int_{-1}^0\!dx\int_x^{0}\!dy\,f(x,y).$$