Question

Find the surface area of that part of the plane 9 x 2 y z = 3 that lies inside the elliptic cylinder \frac{x^2}{81} \frac{y^2}{4} =1 surface area =

Answer 1

To find the surface area of the plane, we must find the partial derivative of
`f(x) = 9x + 2y + z -3`. Then, we have


`\frac {\partial f}{\partial x} = 9`

`\frac {\partial f}{\partial y} = 2`

`\frac {\partial f}{\partial z} = 1`

We can compute for the surface area of the elliptical cylinder,


`<span>(x^2)/(81) + (y^2)/(4) = 1` , with


`Surface Area = \int \int\limits_C { \sqrt{ (9^(2) + 2^(2) + 1^(2))/(1^(2)) } } \, dx \,dy`

knowing that we're working on the plane's surface from the cylinder,


`Surface Area =  √(86) \pi \int \int \,dx  \,dy`

`Surface Area = √(86)  \pi (9-0)(2-0)`

`Surface Area = 18 √(86) \pi`

Thus, the area around the plane of f(x) is
`18 √(86) \pi`.

Answer:
`</span><span>18 √(86) \pi` units²