1. ## Vector Calculus

Calculate the area of the part of the cone $\ z^2 = x^2 + y^2$ that is inside the cylinder $\ x^2 + y^2 \leq 2x$, outside the cylinder $\ x^2 + y^2 \leq 1$ and above the xy plane.

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Cylinder: (ucosv, usinv, u)

Attempt:

2x = 1 ---> x = 1/2 ---> y = $\sqrt(3)/2$

ucos(v) = 1/2
usen(v) = $\sqrt(3)/2$

Solving I get: u = $\sqrt(3)/2$

and

$\int_0^{2\pi}$ $\int_0^{\sqrt{3}/3} u\sqrt{2}dudv$

2. ## Re: Vector Calculus

That would be correct if the base in the xy-plane were the disk with center at (0, 0) with radius $\sqrt{3}/2$ (you have $\sqrt{3}/3$ but you mean $\sqrt{3}/2$). But in fact, it is the intersection of the two disks, $x^2+ y^2\le 1$, the circle with center at (0, 0) and radius 1, and $x^2- 2x+ y^2= (x-1)^2+ y^2= 1$. The points of intersection are $(1/2, \sqrt{3}/2)$ and $(1/2, -\sqrt{3}/2)$ which correspond to $\theta= \pi/3$, r= 1 and $\theta=-\pi/3$, r= 1 in polar coordinates.

You will need to do this as three separate integrals. The main one will be with $\theta$ going from $-\pi/3$ to $\pi/3$, r from 0 to 1. That leaves the two small sections between the straight lines from (0, 0) to $(1/2, \sqrt{3}/2)$ and from (0, 0) to $(1/2, -\sqrt{3}/2)$ and the circle $(x-1)^2+ y^2= 1$. Taking $\theta$ from $\pi/3$ to $\pi/2$, r goes from 1 to the curve $x^2+ y^2= 2x$ which, in polar coordinates is $r^2= 2rcos(\theta)$ or $r(r- 2cos(\theta)= 0$ for r not 0, $r= 2cos(\theta)$ so that r goes from 0 to $2cos(\theta)$. For the last piece, $\theta$ goes from $3\pi/2$ to $2\pi- \pi/3= 5\pi/3$ and, for each $\theta$ r goes from 0 to $2cos(\theta)$ also.

3. ## Re: Vector Calculus

Perfect. Thank you so much for the explanation.