Calculating the area of ​the plane $x+z=1$ portion that is enclosed by $x^2+y^2=2$ with $z≤0$

Question

the exercise is the text in the title. I'm studying surface integrals. To start I thougt to make a change to cartecian coordinates with z as a function of $x$ and $y$, that is, $z=1-x$. With this I make the following parameterization: $O(x,y)=[x,y, x-1]$ Then I calculated the norm: $N=2^{1/2}$ of the normal vector: $[1,0,-1]$ of $O(x,y)$. This is all I did because I do not understand how to project the surface on the xy plane or how to find between the limits of x and y. In addition, how the $z≤0$ condition influences the projection. Another question I have is: is the area of ​​the plane enclosed in the cylinder the same as the area of ​​the portion of the cylinder intersected by the plane?

Answer 1

I don't think the other answer is correct, the constraint that $z\leq 0$ changes things quite a bit.

When you project upwards, this means you are integrating over $$ \{ (x,y):\;x^2+y^2\leq 2\}\cap \{ (x,y):x\geq 1\} $$ Since the combination $z\leq 0$ and $x+z=1$ yields $x\geq 1$.

So as you found, the norm of the unit vector is $\sqrt{2}=||(1,0,1||$ and your integral is $$ \int_{1}^{2}\int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}} \sqrt{2}\;\mathrm dy \mathrm dx $$ Or much more easily in polar coordinates $$ \sqrt{2}\int_1^2 \int_{-\pi/4}^{\pi/4}r\;\mathrm dr\mathrm d\theta\\ =\frac{3\pi \sqrt{2}}{4} $$

Answer 2

It is probably easier not to convert coordinates. To answer your second question: yes the portion enclosed by the cylinder is the same as the area of the intersection.

The intersection between cylinder and the plane is an ellipse. Its semi-minor axis ($a$) is the same as the radius ($R$) of the cylinder. Its semi-major axis ($b$) is $\sqrt2R$ since the plane is at a $45^o$ angle with the base of the cylinder. So we have:

$area\space of \space enclosed\space ellipse=\pi ab = \pi R(\sqrt2R)=\sqrt2\pi R^2=\sqrt2\pi \times2=2\sqrt2\pi.$