Find the area of the plane x-2y + 5z = 13 cut out by the cylinder x² + y²= 9

Answer 1

Invoke the relation dA = dS/√[1 + z_x² + z_y²] between an element (with area dS) of a surface "floating" over the x-y plane and its projection dA onto the "floor" of the x-y plane.

Then the formula for S is ∫_R∫√{1 + (∂z/∂x)² + (∂z/∂y)²}dA; R here is the disk of x² + y² ≤ 9.

From x − 2y + 5z = 13, obtain z = (13 − x + 2y)/5 or 2.6 − 0.2x +0.4y.

Next, ∂z/∂x is ∂(2.6 − 0.2x +0.4y)/∂x or 0 − 0.2 + 0 equal to -0.2.

Then, ∂z/∂y is ∂(2.6 − 0.2x +0.4y)/∂y or 0 − 0 + 0.4 equal to 0.4.

Now, put S together as ∫_R∫√{1 + (∂z/∂x)² + (∂z/∂y)²}dA or ∫_R∫√{1 + (-0.2)² + (0.4)²}dA

which translates here to ∫_R∫√{1 + 0.04 + 0.16}dA or ∫_R∫√{1.2}dA.

This last integrates to √1.2 times A [with A equal to area of R].

From x² + y² = 9, gain A (or area of R) as π(3)² and write the surface area S sought

as 9π • √1.2 or 30.97298093 square units.