Results 1 to 3 of 3

Thread: Vector Calculus

  1. #1
    Newbie
    Joined
    Jun 2011
    Posts
    23

    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.

    ___________________________________________

    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$

    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,769
    Thanks
    3027

    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.
    Last edited by HallsofIvy; Jun 24th 2011 at 03:39 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2011
    Posts
    23

    Re: Vector Calculus

    Perfect. Thank you so much for the explanation.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Vector Calculus (Position Vector)
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Aug 23rd 2011, 01:43 PM
  2. Vector Calculus
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 18th 2011, 10:17 AM
  3. Vector Calculus
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Feb 15th 2011, 01:52 AM
  4. vector calculus - vector feilds
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Feb 25th 2010, 01:17 AM
  5. vector calculus
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Dec 2nd 2009, 02:04 PM

Search Tags


/mathhelpforum @mathhelpforum