(c) How fast is the circular area of the surface of the spherical segment of water growing (in
in.^2 /sec) when the water is 2 inches deep?
The area of the surface of the water is A = πr^2 , where . Thus, A = π(12h − h^2 ).
Taking the derivative of the equation with respect to t, we get
We found in part (a) above that
- Let R be the region in the first quadrant bounded by y^2 = x and x^2 = y.
(a) Find the area of region R.
First, let’s sketch the region.
In order to find the area, we “slice” the region vertically and add up all of the slices. Now, we
use the formula for the area of the region between y = f (x) and y = g(x), from x = a to x = b.
We need to rewrite the equation y^2 = x as y = so that we can integrate with respect to x.
