# Thread: Double integrals

1. ## Double integrals

We are learning about double integrals in 15.5. I think i understand the concepts of the chapter, but i really just don't see how to solve this problem. Any help would be greatly appreciated

2. $\displaystyle x^2+ y^2= a^2$ is, of course, the circle with center at the origin and radius a. The line x+ y= a goes through (a, 0) and (0, a). This problem is ambiguous as there are two regions bounded by those. I am going to assume they the smaller region, to the upper right of the line.
For that region, x obviously varies from 0 to a and, for each a, y varies from y= a- x to $\displaystyle y=\sqrt{a^2- x^2}$. Your integrals should be set up as
$\displaystyle \int_{x=0}^a \int_{y= a- x}^{\sqrt{a^2- x^2}} f(x,y)dydx$.

Do you have to formulas for finding the centroid (in section 15.5)? You should be able to use "symmetry" to cut your work in half.

3. the only equation is for the moment of inertia. i'm not sure that will help in this example though.

4. oh actually there is. the center of mass is given by the second eqn here:

16.5 Applications of Double Integrals.htm

5. Good! In this problem, since the density is constant, say "$\displaystyle \rho$", the mass is just
$\displaystyle M= \rho\int_{x= 0}^a\int_{y= a-x}^{\sqrt{a^2- x^2}} dydx$

and then
$\displaystyle M\overline{x}= \rho\int_{x= 0}^a\int_{y= a-x}^{\sqrt{a^2- x^2}} xdydx$
and
$\displaystyle M\overline{y}= \rho\int_{x= 0}^a\int_{y= a-x}^{\sqrt{a^2- x^2}} ydydx$

The "$\displaystyle \rho$"s will cancel.

6. Accidental double post