a. In this question we are going to derive an integral formula for surface area. Consider a surface 2 = f(x. y) defined over a region R in the xy-plane where f has continuous first partial derivatives. We begin much as we began for volume, by subdiving R into n rectangles where our sides of the rectangles are parallels to the x- and y-axes. Denote a generic little rectangle by Rt. Above this rectangle in the domain there is a little region Si on the surface. The sum of all such little regions Si on the surface will give us an approximation of the total surface area above R. We are going to approximate the area of each little region St by the portion of the tangent plane that lies above Ri, which I will call T. This will look like a parallelogram and we know how to compute the area of a parallelogram if we know what the vectors describing two adjacent sides are by taking the cross product. Your first task is this: use what you know about tangent planes to write vectors at- and bit representing two adjacent sides of the parallelogram that is Ti, in terms of δxi and δyi. Your second task is to use these to compute the area of Ti. Your third task is to put everything into a complete derivation of the formula
Surface Area = ∫∫R√[(∂z/∂x)2 + (∂z/∂y)2 + 1] dA
b. Use the above formula (regardless of whether you properly derived it) to compute the surface area of the portion of the paraboloid z = x2 + y2 below the plane z=1.