Surface Area

    In this topic we apply double integrals to the problem of computing the area of a surface over a region. We demonstrate a formula that is analogous to the formula for finding the arc length of a one variable function and detail how to evaluate a double integral to compute the surface area of the graph of a differentiable function of two variables.

Proposition (Surface Area) Assume that the function has continuous partial derivatives and in a region of the -plane. Then the portion of the surface that lies over has surface area



    Proof. Consider a surface defined by defined over a region of the Enclose the region in a rectangle partitioned by a grid with lines parallel to the coordinate axes. This creates a number of cells, and we let ...., denote those that lie entirely within For let be any corner of the rectangle and let be the tangent plane above on the surface of Let denote the area of the patch of the surface that lies directly above The rectangle projects onto a parallelogram in the tangent plane and if is small we would expect the area of this parallelogram to approximate closely the element of the area If and are the lengths of the sides of the rectangle the approximating parallelogram will have sides determined by the vectors



and



If is the area of the approximating parallelogram, we have



To compute we first find the cross product:





Then we calculate the norm,





Finally, summing over the entire partition, we see that the surface area over may be approximated by the sum



where   This is a Riemann sum, and by taking an appropriate limit (as the partition becomes more refined), we find that the surface area satisfies

Example (Surface Area) Find the surface area.

(a) Find the surface area of the part of the surface that lies above the triangular region in the with vertices and

    Solution. The region is described by



We have











(b) Find the surface area of the part of the paraboloid that lies under the plane
    
    Solution. The plane intersects the paraboloid in the circle Therefore, the given surface lies above the disk with center the origin and radius 3. Converting to polar coordinates, we have,

    It is worth noting the following similarities.

Example (Computing Surface Area) Find the surface area of the portion of the sphere that lies inside the cylinder
    
    Solution. Let Then  




and



The projected region in polar form is Since half the surface is above the and half the surface is below, we have

Example (Surface Area of a Sphere) Find the surface area of a sphere of radius   

    Solution. Let  



Then





and



In polar form, we have

Example (Surface Area of a Cylinder) Find the surface area of a cylinder of radius and height
    
    Solution. Let



Then





and



We have,

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