omg this hw problem is making me go crazy please hlep?

Question

http://img45.imageshack.us/img45/6987/centroidqn7.jpg
Determine by direct integration the centoid of area shown\
Okay what I did was I figured it was a quarter-semicircular minus a triangle. I start off by getting the area for the quarter shape intefral of y = (b*sqrt(1-(x^2/a^2)), then I get the X coordinate by doing the integral of xy and diving it by the area. Im stuck on the area and coordinates for the second shape tho what do i use??? the formula for the are is Integral of ydx but I aleady used it for the first shape, i dont udnertand please help =(

Answer 1

The curve is an ellipse with semi-major axis = a and semi- minor axis = b. The area under cosideration is simply that part of an ellipse that has a center at the origin and lies in the first quadrant.

So compute the area of 1/4 of an ellipse by whatever method you want. Then compute First moment w/r x and w/r to y.

Then compute xbar and ybar.

Answer 2

You will notice u can just do 1/4 the area of the circle minus the area of the triangle. However, since the directions tell you to do direct integration...

Since you are integrating w.r.t. x, your limits should be from 0 to a for the circular part. Then your triangle can be integrated within the same limits, with the integrand as the equation of that linear line (y = (-b/a)x+b)).

The answer should be like (.25)pi((b+a)/2)^2 + (b*a)/2 or something of that nature.