Find the area of the part of the plane $x+y+z=a$ limited by the cylinder $x^2+y^2=b^2$

Question

Find the area of the part of the plane $x+y+z=a$ limited by the cylinder $x^2+y^2=b^2$

I'm having issues on how to parametrise the surface, I've tried with $\vec{r}(b,\theta)=(b\cos\theta,b\sin\theta, a-b^2)$ since it "fits" both equations perfectly, but I don't get to the right answer so I'm doubtious about that part. After parametrising it should be easy as I know the surface is $$S=\iint_S ||\vec{N}(b,\theta)||dbd\theta$$ being $\vec{N}=\frac{d\vec{r}}{db}\times \frac{d\vec{r}}{d\theta}$. So what I'm really having troube is parametrising the surface.

Answer 1

Since the surface has support:

$$ \color{red}{\Sigma} := \left\{(x,y,z) \in \mathbb{R}^3 : x+y+z=a, \; x^2+y^2 \le b^2\right\} $$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad$

then a natural parameterization is the following:

$$ \mathbf{r}(u,v) = \left(u\cos v,\,u\sin v,\,a-u\cos v-u\sin v\right) \quad \quad \text{with} \; (u,v) \in D := [0,b] \times [0,2\pi) $$

from which, simply:

$$ ||\color{red}{\Sigma}|| := \iint\limits_{\color{red}{\Sigma}} 1\,\text{d}S = \iint\limits_D ||\mathbf{r}_u(u,v) \land \mathbf{r}_v(u,v)||\,\text{d}u\,\text{d}v = \int_0^{2\pi} \text{d}v \int_0^b \sqrt{3}\,u\,\text{d}u = \boxed{\sqrt{3}\,\pi\,b^2}. $$

---

Equivalently, it could have been observed that:

$$ z = f(x,y) = a-x-y \quad \quad \Rightarrow \quad \quad ||\nabla f||^2 = (-1)^2 + (-1)^2 = \color{magenta}{2} $$

from which:

$$ ||\color{red}{\Sigma}|| := \iint\limits_{\color{red}{\Sigma}} 1\,\text{d}S = \iint\limits_{x^2+y^2 \le b^2} \sqrt{1+\color{magenta}{2}}\,\text{d}x\,\text{d}y = \sqrt{3}\iint\limits_{x^2+y^2 \le b^2} 1\,\text{d}x\,\text{d}y = \boxed{\sqrt{3}\,\pi\,b^2}. $$

Answer 2

Note that the cutout of the plane by the cylinder is an ellipse with semi-minor axis $b$. To find the semi-major axis, calculate the angle $\theta$ between the normal vector of the plane $(1,1,1)$ and the $z$-axis $$\cos\theta = \frac{(0,0,1)\cdot(1,1,1)}{\sqrt3}=\frac1{\sqrt3}$$ Thus, the semi-major axis is $\frac b{\cos\theta}=\sqrt3b$ and the area of the ellipse is $$A= \pi\cdot b\cdot \sqrt3b=\sqrt3\pi b^2$$