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For the following question:

$\displaystyle \int\!\!\!\int_D x^2y\,dx\,dy$ where $\displaystyle D$ is the bounded region $\displaystyle x= y^2$ & $\displaystyle y = \frac{1}{2}x$

I get the limits of integration to be:

$$\int_0^4\!\!\!\int_{2y}^{y^2} x^2y\,dx\,dy$$

I got the limts for y by substituting $\displaystyle y = \frac{1}{2}x$ in to $x= y^2$

Is this the right approach (probably not)?

I'm sure my graph is ok, brain freeze from here though.

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I guess you are asking for the integral over the region bounded by the line and the parabola. – user20266 Apr 10 '12 at 16:33
What makes you think you're wrong? – nbubis Apr 10 '12 at 16:36
up vote 4 down vote accepted

For double integration problems, it's always beneficial to draw a diagram of the region of integration when you can:

enter image description here

The limits on the inner integral (it's a $dx$ integral, so horizontal in nature) correspond to integrating over the blue line segment in the diagram for a fixed $y$ value (you were almost correct here, but got the order wrong, integrate from the leftmost point ($x=y^2$) to the rightmost point ($x=2y$) ). The limits on the outer integral correspond to the $y$-range the horizontal line segments "sweep through" (from $y=0$ to $y=2$, here).

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I'm interested in what tool you use to draw the diagram. – Jack Apr 10 '12 at 17:58
@Jack The Javascript library JSXgraph. – David Mitra Apr 10 '12 at 18:04
Yep thanks David. I was close but no cigar. I drew the graph, but should have evaluated the numbers a bit better... I suspect I may have been half asleep or something because the inner integral is obvious when I look at it now. Cheers – Billy Ray Valentine Apr 10 '12 at 19:25

The points of intersection are $(x,y)=(0,0)$ and $(x,y)=(4,2)$. First you have $$ \int_\text{?}^\text{?}\cdots\cdots\cdots dy. $$ The variable $y$ goes from $0$ to $2$.

Then you have $$ \int_0^2\left(\int_\text{?}^\text{?}\cdots\cdots\,dx\right) \, dy. $$ For any particular value of $y$, the other variable $x$ goes from $y^2$ up to $2y$.

Notice that $y^2$ is less than $2y$ if $y$ is between $0$ and $2$.

So you've got $$ \int_0^2\left(\int_{y^2}^{2y}\cdots\cdots\,dx\right) \, dy. $$

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Really helpful dude. Thanks. This and the answer below illustrate where I was going wrong. – Billy Ray Valentine Apr 10 '12 at 19:25

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