1. ## Vector Calculus

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

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Cylinder: (ucosv, usinv, u)

Attempt:

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

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

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

and

$\displaystyle \int_0^{2\pi}$$\displaystyle \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 $\displaystyle \sqrt{3}/2$ (you have $\displaystyle \sqrt{3}/3$ but you mean $\displaystyle \sqrt{3}/2$). But in fact, it is the intersection of the two disks, $\displaystyle x^2+ y^2\le 1$, the circle with center at (0, 0) and radius 1, and $\displaystyle x^2- 2x+ y^2= (x-1)^2+ y^2= 1$. The points of intersection are $\displaystyle (1/2, \sqrt{3}/2)$ and $\displaystyle (1/2, -\sqrt{3}/2)$ which correspond to $\displaystyle \theta= \pi/3$, r= 1 and $\displaystyle \theta=-\pi/3$, r= 1 in polar coordinates.

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

3. ## Re: Vector Calculus

Perfect. Thank you so much for the explanation.