Double Integral Limits

**algorithm** · June 1st 2009, 04:03 PM

Hi,

I don't understand how the limits in the solution of part (b) have been obtained. Please explain how it's done.

Thanks

**Danny** · June 1st 2009, 04:14 PM

Originally Posted by algorithm

Hi,

I don't understand how the limits in the solution of part (b) have been obtained. Please explain how it's done.

Thanks

$\int_{y=a}^{y=b} \int_{x=f(y)}^{x=g(y)} x^2+y^2 \,dx\, dy$

where $x = f(y)$ the left curve of your region, $x = g(y)$ the right curve of your region, $y = a$ the bottom point of your region and $y = b$ , the top point of your region.

**Danneedshelp** · June 1st 2009, 04:47 PM

Well, the limits from the first integral are given by the line $y=\frac{x}{\sqrt{3}}$ and the section of the circle equation given by $x^{2}+y^{2}=1^{2}$ .

First, we can see that the first (inner) most integral is with respect to $x$ ; thus, we must solve our two equations for $x$ . Also, notice the second integral is with respect to $y$ . Therefore, we must make sure we only have y's in our second integrand. Solving our two equations for $x$ will give us that.

So, solving for $x$ we get $x=y\sqrt{3}$ and $x=\sqrt{1-y^{2}}$ .

Therefore, we have $\int_{y\sqrt{3}}^{\sqrt{1-y^{2}}}(x^2+y^2) \,dx$ as our inside integral. The limits are ordered the way they are, because the circle is above the line.

The outside integral just tells us how far up the y-axis to integrate. So, that’s where the limits are coming from.

And, just a note, when you are dealing with iterated integrals, always make sure your outer most limits are numbers and not functions.

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