The area enclosed by the curve $y^2+x=0$ and the line $y=x+2$ has a mass equal to $9/2$ (taking density to be uniform). How do I find the center of gravity of this area?
My problem arises in finding the first moment wrt x-axis. I'm sure the answer is obtained as the addition of two integrals, but I'm having difficulty in finding the appropriate limits of integration.
Best Answer
The integral you want is $\int_{-2}^{1}\int_{y-2}^{y^2}....dxdy$, where ... is the integrand needed for center of gravity.
You can use the same limits for both. The integral for $\bar{y}$ will be $\int_{-2}^{1} y(y^2-y+2)dy$
I believe I made a mistake! The upper limit on x should be$-y^2$, not $y^2$ in both cases. However for $\bar{x}$ it doesn't matter!
$\bar{y}=\int_{-2}^{1}y(2-y-y^2)dy$.
Further error, I forgot to divide by the area. Net result $\bar{y}=\frac{-9}{16}$.