Answer:
Integrate( sqrt(9-x^2) from x=-3 to x=3)
Explanation:
The equation for a full circle is (x-h)^2+(y-k)^2=r^2 where (h,k) is center and radius is r.
Your center, your (h,k) is (0,0). Your radius, your r, is 3.
So your equation is (x-0)^2+(y-0)^2=3^2 or more simply x^2+y^2=9.
We also must consider we don't have full circle.
Solving for y will give us the circle in terms of top half if we take positive values and bottom half if we take negative values. Since y is positive in the picture, you only see top half, we will only take the positive cases for y.
Subtracting x^2 on both sides gives: y^2=9-x^2
Square root both sides: y= sqrt(9-x^2)
(I did not choose -sqrt(9-x^2) because again y is positive).
So the x's in the picture range from -3 to 3.
The integral is therefore,
Integrate( sqrt(9-x^2) from x=-3 to x=3)