Find surface area of $z=x+3$ with $x^2+y^2\leq 1$

Question

Find the surface area of $z=x+3$ with $\{(x,y,z)\mid x^2+y^2\leq 1\}$

So we first look at the projection of $\phi(x,y)=(x,y,x+3)$ on $xy$

Then area element is $\sqrt{1+f_x^2+f_y^2}=\sqrt{1+1^2+0^2}=\sqrt{2}$

$$\sqrt{2}\int\int dx\,dy$$

$x=r \cos t,y=r \sin t$

So $$\sqrt{2}\int_{0}^{1}\int_{0}^{2\pi} r \,dt\,dr=\sqrt{2}\pi$$

Is it correct? can we use Green/Stokes to solve it?

Answer 1

What you have is correct.

I am not sure why you want to use Stokes... but supposing you did...you could say that the contour of your region is bound by the contour.

$(x,y,z) = r(t) = (\cos t, \sin t, \cos t+3)\\ r'(t) = (-\sin t, \cos t, -\sin t)$

Now we need a function $F(x,y,z)$ such that $\nabla \times F$ is perpendicular to the plane and of magnitude 1.

$F = (0,\frac {\sqrt2}{2}(x+z), 0)$ would suffice.

$\int_0^{2\pi} F\cdot dr = \sqrt 2\pi$

Answer 2

Yes it is correct, indeed the intersection is the area enclosed by an ellipse with axes $a=2$ and $b=2\sqrt 2$ that is

$$A=\frac14 \pi \cdot 2 \cdot 2\sqrt 2=\sqrt 2 \pi$$