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MI.6 Surface Area

In this section we apply double integrals to the problem of computing the area of a surface. To do this we let S be a surface with equation , where is continuous partial derivatives. For simplicity in deriving the surface area formula, we assume that $\geq 0$ and the domain of is a rectangle. We divide D into small rectangles R $_{ij}$ with area If $(x_{i},y_{i})$ is the corner of R $_{ij}$ closest to the origin, let P be the point on S directly above it. The tangent plane to S at P $_{ij}$ is an approximation to S near P $_{ij}.$ So the area $\Delta T_{ij}$ of the part of this tangent plane that lies directly above R $_{ij}$ is an approximation to the area $\Delta S_{ij}$ of the part of S that lies directly above R $_{ij}.$ Thus, the sum is an approximation to the total area of S, and this approximation appears to improve as the number of rectangles increases. Therefore, we define the surface area of to be:

Using the definition of a double integral and a little algebra, we come upon the following definition for the surface area:

Also, an alternative notation:

Also, notice the similarity of the surface area formula with the arc length formula:

Example

Find the surface area of the part of the paraboloid $z=x^{2}+y^{2}$ that lies under the plane

Solution:

The plane intersects the paraboloid in the circle $x^{2}+y^{2}=9,~z=9$ (Like a level curve) Therefore the given surface lies above the disk D with center the origin and radius 3.

picture of $z=x^{2}+y^{2}$ and the plane z=9. with the disk on the xy plane highlighted and with the function above that disk highlighted

Using the surface area formula we have:

For ease of integration, this can be converted into polar coordinates:

Now we simply integrate:

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