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

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

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

We have

$\dfrac{\partial f}{\partial x} = -\dfrac13$

$\dfrac{\partial f}{\partial y} = -\dfrac23$

so that the area element is

$dA = \sqrt{1 + \left(-\dfrac13\right)^2 + \left(-\dfrac23\right)^2} \,dx\,dy = \dfrac{\sqrt{14}}3\,dx\,dy$

Then we have

$\displaystyle \iint_R dA = \frac{\sqrt{14}}3 \iint_R dx \, dy$

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

$\displaystyle \iint_R dA = \frac{\sqrt{14}}3 \cdot 6\pi = \boxed{2\sqrt{14}\pi}$