Results 1 to 4 of 4

Thread: Calculate the surface area enclosed by y=4-x and the x axis between 0≤ x ≤2?

  1. #1
    Newbie
    Joined
    Jul 2014
    From
    scotland
    Posts
    1

    Calculate the surface area enclosed by y=4-x and the x axis between 0≤ x ≤2?

    Calculus assignment question
    Centroid?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,320
    Thanks
    1289

    Re: Calculate the surface area enclosed by y=4-x and the x axis between 0≤ x ≤2?

    Are you calculating the area bounded by the curves $\displaystyle y=4-x^2, y=0, x=0$? Surface area could imply the surface area of a volume (obtained by revolution), and would greatly change the answer.

    Assuming you are just looking for bounded area, what do you mean by "centroid"? The area between the x-axis and the curve $\displaystyle y=4-x^2$ is just $\displaystyle \int_0^2(4-x^2)dx$. That is the very definition of area under the curve.

    Edit: Oh, by centroid, do you mean the "center" of that area? Let $\displaystyle k$ be a number such that $\displaystyle \int_0^k(4-x^2)dx = \int_k^2(4-x^2)dx$ and $\displaystyle 0<k<2$. Then the centroid is $\displaystyle \left(k,2-\dfrac{k^2}{2}\right)$. You probably don't have to solve the resulting cubic, but if you did, you could show $\displaystyle k = 4\sin\left(\dfrac{\pi}{18}\right)$
    Last edited by SlipEternal; Jul 3rd 2014 at 08:59 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,320
    Thanks
    1289

    Re: Calculate the surface area enclosed by y=4-x and the x axis between 0≤ x ≤2?

    I calculated the y-component of the centroid wrong. I think it would actually be some value $\displaystyle p$ such that $\displaystyle \int_0^p\sqrt{4-y}dy = \int_p^2\sqrt{4-y}dy$.

    So, the actual centroid would be $\displaystyle \left(4\sin\left(\frac{\pi}{18}\right),4-\sqrt[3]{18+8\sqrt{2}}\right)$
    Last edited by SlipEternal; Jul 3rd 2014 at 10:33 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,718
    Thanks
    3003

    Re: Calculate the surface area enclosed by y=4-x and the x axis between 0≤ x ≤2?

    There is NO "surface area" bounded by $\displaystyle y= 4- x^2$ and the x-axis! I suspect you mean the surface area of the figure generated by rotating $\displaystyle y= 4- x^2$ around the x-axis, with x between 0 and 2 (so y between 0 and 4).

    We can write parametric equations for that surface by taking y and z to be $\displaystyle y= r cos(\theta)$ and $\displaystyle z= r sin(\theta)$ so that $\displaystyle x= \sqrt{4- y}= \sqrt{4- r cos(\theta)$ with r and $\displaystyle \theta$ as parameters, going from 0 to 4 and $\displaystyle \theta$ from 0 to $\displaystyle 2\pi$. We can then write a point on that surface as a vector equation $\displaystyle \vec{p}= \sqrt{4- rcos(\theta)}\vec{i}+ r cos(\theta)\vec{j}+ r sin(\theta)\vec{k}$.

    Now the derivatives, $\displaystyle \vec{p}_r= \frac{1}{2}(4- r cos(\theta)^{-1/2}(- cos(\theta))\vec{i}+ cos(\theta)\vec{j}+ sin(\theta)\vec{k}$ and $\displaystyle \vec{p}_\theta= \frac{1}{2}(4- r cos(\theta)^{-1/2}(r sin(\theta)\vec{i}- r sin(\theta)\vec{j}+ r cos(\theta)\vec{k}$ are tangent to the surface. Their cross product, $\displaystyle \frac{1}{2}r(4- r cos(\theta))\vec{j}$ has length $\displaystyle \frac{r}{2}(4- r cos(\theta))$ so that the "integral of surface area is $\displaystyle \frac{r}{2}(4- rcos(\theta))dr d\theta$.

    The surface area is given by $\displaystyle \frac{1}{2}\int_{\theta= 0}^{2\pi}\int_{r= 0}^4 r(4- r cos(\theta))dr d\theta$
    Last edited by HallsofIvy; Jul 3rd 2014 at 11:27 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Mar 27th 2012, 10:34 PM
  2. Replies: 7
    Last Post: Mar 18th 2012, 01:11 PM
  3. Replies: 2
    Last Post: Mar 19th 2011, 01:38 AM
  4. Replies: 3
    Last Post: Jul 27th 2010, 03:46 PM
  5. Replies: 1
    Last Post: May 27th 2008, 07:56 AM

Search Tags


/mathhelpforum @mathhelpforum