Graph the integrand and use area to evaluate the definite integral from -9 to 9 of the square root of an 81 -, X ^2 DX. So the first thing we want to realize is that the function is sqrt 81 -, X ^2. So if we thought about calling that Y, we can see that the Y is always going to be greater than or equal to 0. So I'm going to square each side and get y ^2 equaling 81 -, X ^2. Add the X ^2 to the other side and this is just an equation for a circle where my Y has to be positive. So X ^2 + y ^2 equal 81. So the center is at 00 and the radius is 9. Now the integrand is going from -9 to 9 for our X values. So we want the top half of a circle. So the area is going to be 1/2 Pi R-squared and in this case our R is 9.