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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?

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First you integrate wrt to x, so in the limits you have x = -1 and x = y. For y, you have y = -1 and y = 0. You should also try accepting your answers for your other questions if you want help –  sidht Aug 15 '12 at 18:59

3 Answers 3

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 $$\int_{-1}^{0} \!dy \int_{-1}^{y} \!dx\, f(x,y) = \int_{-1}^0\!dx\int_x^{0}\!dy\,f(x,y).$$

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YES.+1. $ $ $ $ –  Did Aug 15 '12 at 22:12
    
+1, of course. :) –  J. M. Aug 15 '12 at 23:16

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.

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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 $-1 < x < 0$.

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