# Thread: Calculate the surface area enclosed by y=4-x² and the x axis between 0≤ x ≤2?

1. ## Calculate the surface area enclosed by y=4-x² and the x axis between 0≤ x ≤2?

Calculus assignment question
Centroid?

2. ## Re: Calculate the surface area enclosed by y=4-x² 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. 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 $k = 4\sin\left(\dfrac{\pi}{18}\right)$

3. ## Re: Calculate the surface area enclosed by y=4-x² 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 $\left(4\sin\left(\frac{\pi}{18}\right),4-\sqrt[3]{18+8\sqrt{2}}\right)$

4. ## Re: Calculate the surface area enclosed by y=4-x² 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 $\frac{1}{2}\int_{\theta= 0}^{2\pi}\int_{r= 0}^4 r(4- r cos(\theta))dr d\theta$