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$z=\sqrt{x^2+y^2}$ is the surface we working on.

I am a bit stuck on choosing the limits for this problem, I have done the following:

$J(\text{jacobian})=\sqrt{Z^2_x+Z^2_y+1}=\left(\dfrac{x^2+y^2}{(\sqrt{x^2+y^2})^2}+1\right)^{\frac{1}{2}}=\sqrt{2}$ $$\iint_S f \, dS=\iint \sqrt{2} \, dz \, d\theta.$$

I can't explain why I chose $dz, d\theta$, I guess since the one surface is bounded by $z$ and $x^2+y^2=x$ looks like the equation of a circle so $\theta$ :)

Anyway, I know that $z \left[0:\sqrt{x^2+y^2} \right] \rightarrow \left[0:r \right]$

and

$x^2+y^2=x\rightarrow r^2=r\cos(\theta)$, I tried to solve for $\theta$ from that but no luck. Please assist -Thanks.

I looked at this post here and it doesn't really address my issue.

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HappyFeet
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1 Answers1

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No need to solve for $\theta$. Notice that $z$ is not an independent variable, thus not a good choice. Using $x = r \cos(\theta)$ and $y = r \sin(\theta)$ you obtain $z=r$. The length of the normal vector is $$\sqrt{ \left( \frac{\partial z}{\partial r} \right)^2 + \left( \frac{\partial z}{\partial \theta} \right)^2 + 1 } = \sqrt{2}.$$ The surface area is $$\iint_S dS = \iint_S \| {\mathbf r}_u \times {\mathbf r}_v \| \, dA.$$ Notice that this is not a Jacobian. Jacobians arise in changes of variable of the form $\mathbb{R}^2 \to \mathbb{R}^2$ or $\mathbb{R}^3 \to \mathbb{R}^3$. Since this is the graph of a function, the natural parameterization is $${\mathbf r} = (r \cos(\theta), r \sin(\theta), r).$$ The limits for $\theta$ are $0 \leq \theta \leq 2 \pi$ and for $r$ are $0 \leq r \leq \cos(\theta)$. Your area integral becomes $$\iint dS = \int_0^{2 \pi} \int_0^{\cos(\theta)} \sqrt{2} r \, dr \,d \theta.$$

Mark Fantini
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  • $\theta=0 \rightarrow 2\pi$ because we have a full circle correct? – HappyFeet Dec 30 '14 at 10:01
  • @Ozwurld Yes, that's correct. I've added more explanations to account for misconceptions in your post. – Mark Fantini Dec 30 '14 at 10:03
  • So should i always parametrize using independent variables?, and can i use $${\mathbf r} = (r \cos(\theta), r \sin(\theta)).$$ for functions that do not contain circular regions? – HappyFeet Dec 30 '14 at 10:08
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    @Ozwurld You can, it just may not be useful. And yes, the number of independent variables has to do with the dimension of the object you are dealing with. In this case the easiest method we could use to solve the problem involved letting $z$ be a dependent variable. Generally polar coordinates are only useful for objects with cylindrical symmetry. – Mark Fantini Dec 30 '14 at 11:16