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

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

    Calculus assignment question
    Centroid?
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    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 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 y=4-x^2 is just \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 k be a number such that \int_0^k(4-x^2)dx = \int_k^2(4-x^2)dx and 0<k<2. Then the centroid is \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 k = 4\sin\left(\dfrac{\pi}{18}\right)
    Last edited by SlipEternal; July 3rd 2014 at 09:59 AM.
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    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 p such that \int_0^p\sqrt{4-y}dy = \int_p^2\sqrt{4-y}dy.

    So, the actual centroid would be \left(4\sin\left(\frac{\pi}{18}\right),4-\sqrt[3]{18+8\sqrt{2}}\right)
    Last edited by SlipEternal; July 3rd 2014 at 11:33 AM.
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    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 y= 4- x^2 and the x-axis! I suspect you mean the surface area of the figure generated by rotating 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 y= r cos(\theta) and z= r sin(\theta) so that x= \sqrt{4- y}= \sqrt{4- r cos(\theta) with r and \theta as parameters, going from 0 to 4 and \theta from 0 to 2\pi. We can then write a point on that surface as a vector equation \vec{p}= \sqrt{4- rcos(\theta)}\vec{i}+ r cos(\theta)\vec{j}+ r sin(\theta)\vec{k}.

    Now the derivatives, \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 \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, \frac{1}{2}r(4- r cos(\theta))\vec{j} has length \frac{r}{2}(4- r cos(\theta)) so that the "integral of surface area is \frac{r}{2}(4- rcos(\theta))dr d\theta.

    The surface area is given by \frac{1}{2}\int_{\theta= 0}^{2\pi}\int_{r= 0}^4 r(4- r cos(\theta))dr d\theta
    Last edited by HallsofIvy; July 3rd 2014 at 12:27 PM.
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