1. ## Double Integral Limits

Hi,

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

Thanks

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

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

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

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

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

Therefore, we have $\displaystyle \int_{y\sqrt{3}}^{\sqrt{1-y^{2}}}(x^2+y^2) \,dx$ 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.