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?
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?
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 $\int_{-1}^{0} \!dy \int_{-1}^{y} \!dx\, f(x,y) = \int_{-1}^0\!dx\int_x^{0}\!dy\,f(x,y).$
If you draw it, you will see that the domain of integration is a triangle with vertices $(-1,-1)$, $(-1,0)$ and $(0,0)$. You should be able to find the new limits of integration from this.
We know that $dx$ ranges from $-1$ to the line $x = y$. Look at the following graph
http://www.wolframalpha.com/input/?i=plot+y+%3D+x
This means that every infinitesimal element $x$ starts from the $-1$ position along the $x$ axis and goes to the line $y = x$ (horizontally). Next we know $dy$ ranges from $-1$ to $0$, so every infinitesimal element $y$ starts at $-1$ along the $y$ axis and goes up to zero (vertically). Essentially forming a triangle in the 3rd quadrant below the $x$ axis and above the line $y = x$.
Now instead of drawing small $dx$ first you want to draw small $dy$ first when you switch the order. So given the same triangle we know that a small $dy$ will start at the line $y = x$ and go up to zero. Giving the bounds $x < y < 0$, and we want $x$ to range from $-1$ to $0$ giving the bounds $-1 < x < 0$.