Question

Find the area of the surface.the part of the plane x 2y 3z = 1 that lies inside the cylinder x2 y2 = 6

Answer 1

I assume you mean the plane
`x+2y+3z=1`. Its area over the region

`R = \left\{(x,y) ~:~ x^2 + y^2 \le 6\right\}`

is given by the integral

`\displaystyle \iint_R dA = \iint_R \sqrt{1 + \left((\partial f)/(\partial x)\right)^2 + \left((\partial f)/(\partial y)\right)^2} \, dx \, dy`

where
`z=f(x,y) = \frac{1 - x - 2y}3`.

We have

`(\partial f)/(\partial x) = -\frac13`

`(\partial f)/(\partial y) = -\frac23`

so that the area element is

`dA = √(1 + \left(-\frac13\right)^2 + \left(-\frac23\right)^2) \,dx\,dy = \frac{√(14)}3\,dx\,dy`

Then we have

`\displaystyle \iint_R dA = \frac{√(14)}3 \iint_R dx \, dy`

and the remaining integral is exactly the area of the disk
`x^2+y^2\le6`. Its radius is √6, so its area is π (√6)² = 6π. So the area of the surface is

`\displaystyle \iint_R dA = \frac{√(14)}3 \cdot 6\pi = \boxed{2√(14)\pi}`