Question

Find the area of the part of the plane z=ax+by+c that projects onto a region in the xy-plane with area A.

Answer 1

The area of the part of the plane z = ax + by + c that projects onto a region in the xy-plane with area A is $\sqrt{1+a^2+b^2} A$.

What is the formula for calculating the area of the part of a plane surface?

The formula for calculating the surface area of the required part of the plane when projected onto a region in the xy-plane is

$\int\int_D\sqrt{1+(\frac{\partial z}{\partial y})^2+(\frac{\partial z}{\partial x})^2 } dA$

Where D is the projection of the surface on the xy-plane with area A.

Calculation:

The given plane is z = ax + by + c

Calculating the partial differentiation of z w.r.t x and y:

$\frac{\partial z}{\partial x}=\frac{\partial}{\partial x}(ax + by + c)$

    = a

Similarly,

$\frac{\partial z}{\partial y}=\frac{\partial}{\partial y}(ax + by + c)$

    = b

Then, calculating the surface area,

A(S) = $\int\int_D\sqrt{1+(\frac{\partial z}{\partial y})^2+(\frac{\partial z}{\partial x})^2 } dA$

       = $\int\int_D\sqrt{1+(b)^2+(a)^2 } dA$

       = $\sqrt{1+a^2+b^2}\int\int_DdA$

Since the area of the xy-plane is A, the integration over dA becomes A.

Thus,

Surface area = $\sqrt{1+a^2+b^2}A$

Learn more about the area of the plane when projected onto xy-plane here:

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