Surface Integral: dot product of two unit vectors

[Matty R]

Hello.

I understand most of the work involved with these types of questions, but there is one point in an example I'm following that I don't understand.

Homework Statement

Evaluate:

$$I = \int{(z^2)}dS$$ over the positive quadrant of a sphere, where (x,y > 0).

Homework Equations

$$x^2 + y^2 + z^2 = 1$$
$$\underline{\hat{n}} = \frac{\nabla f}{|\underline{\nabla}f|}$$

The Attempt at a Solution

Project onto the xy plane:

$$dS = \frac{dxdy}{\underline{\hat{n}} \cdot \underline{\hat{k}}}$$

At any point on the surface:

$$z^2 = 1 - x^2 - y^2$$

Therfore:

$$\int{(z^2)}dS = \int{\int{(1-x^2-y^2)}}\frac{dxdy}{\underline{\hat{n}} \cdot \underline{\hat{k}}}$$

$$\underline{\hat{n}} = \frac{\nabla f}{|\underline{\nabla}f|}$$

$$= x \underline{\hat{i}}+y\underline{\hat{j}}+\left (\sqrt{1-x^2-y^2} \right)\underline{\hat{k}}$$

I don't understand how to get to the next step:

$$\underline{\hat{n}} \cdot \underline{\hat{k}} = \sqrt{1-x^2-y^2}$$

Its probably something really simple, knowing me.

I've tried the dot product, but couldn't get the answer from that.

Would anyone be gracious enought to end my torment?

I hope my Tex is okay. Its changed since I was last here, though it could be my reinstallation of ProText. Everything is in bold and the gaps between equations are much bigger.

Thanks.

 

[HallsofIvy]

Personally, I wouldn't do it that way. I particularly dislike the notation
$$\vec{n}= \frac{\nabla f}{|\nabla f}$$
because you also have a factor of $|\nabla f|$ in dS itself- they always cancel and so there is no need to calculate it.

Instead, I think of a surface integral this way: parametric coordinates for the surface of a sphere, of radius R can be got from the formulas for spherical coordinates with $\rho$ fixed at R. That is, $x= Rcos(\theta)sin(\phi)$, $y= Rsin(\theta)sin(\phi)$, $z= Rcos(\phi)$. That is, we can write the "position vector" of any point on the surface of the sphere as
$$\vec{r}(\theta, \phi)= Rcos(\theta)sin(\phi)\vec{i}+ Rsin(\theta)sin(\phi)\vec{j}+ Rcos(\phi)\vec{k}$$

The derivatives with respect to $\theta$ and $\phi$,
$$\vec{r}_\theta= -Rsin(\theta)sin(\phi)\vec{i}+ Rcos(\theta)sin(\phi)\vec{j}$$
$$\vec{r}_\phi= Rcos(\theta)cos(\phi)\vec{i}+ Rsin(\theta)cos(\phi)\vec{j}- Rsin(\phi)\vec{k}$$
are vectors lying in the tangent plane to the sphere. Their cross product,
$$R^2cos(\theta)sin^2(\phi)\vec{i}+ R^2sin(\theta)sin^2(\phi)\vec{j}+ R^2 sin(\phi)cos(\phi)\vec{k}$$
where the order of multiplicatio/n has been chosen to give the outward pointing normal, is normal to the sphere and, with $d\theta d\phi$, gives the "vector differential of surface area".

dS itself, the "scalar differential of surface area" is given by the magnitude of that vector:
$dS= R^2 sin(\phi)d\theta d\phi$.

Your integrand is $z^2= R^2cos^2(\phi)$ so your integral is
$$R^4 \int\int cos^2(\phi)sin(\phi)d\phi d\theta$$
where the integral is to include only the first octant.

(Of course, in this problem, R= 1.)

 

[HallsofIvy]

Having got that off my chest, you can, of course, to answer your actual question, yes, it is "something really simple"!

You arrived at the fact that $\vec{n}= x\vec{i}+ y\vec{j}+ \sqrt{1- x^2- y^2}\vec{k}$.

It's dot product with $\vec{k}= 0\vec{i}+ 0\vec{j}+ 1\vec{k}$ is
$$0(x)+ 0(y)+ 1(\sqrt{1- x^2- y^2})= \sqrt{1- x^2- y^2}$$

 

[Matty R]

Wow. Thanks for the replies.

I knew it would be something straightforwards like that. I keep overthinking.

I completely agree about the notation. The method I've been using looks very messy, so I'll give the method you prefer a go.

Just past the bit where I was stuck, the example has this:

$$\int{(z^2)dS} = \int{\int{(1-x^2-y^2)\frac{dxdy}{\sqrt{1-x^2-y^2}}}}$$

$$= \int{\int{(1-x^2-y^2)}dxdy}$$

Is that what you meant about the cancelling terms? I don't understand it, though it looks to be saying z = 1, or the dot product of the two unit vectors is 1.

 

[HallsofIvy]

Wow. Thanks for the replies.

I knew it would be something straightforwards like that. I keep overthinking.

I completely agree about the notation. The method I've been using looks very messy, so I'll give the method you prefer a go.

Just past the bit where I was stuck, the example has this:

$$\int{(z^2)dS} = \int{\int{(1-x^2-y^2)\frac{dxdy}{\sqrt{1-x^2-y^2}}}}$$

$$= \int{\int{(1-x^2-y^2)}dxdy}$$

What? No,
$$\frac{1- x^2- y^2}{\sqrt{1- x^2- y^2}}= \sqrt{1- x^2- y^2}$$
not just $1- x^2- y^2$. You will need a couple of trig substitutions to integrat that.

[tex]Is that what you meant about the cancelling terms? I don't understand it, though it looks to be saying z = 1, or the dot product of the two unit vectors is 1.[/QUOTE]

 

[Matty R]

Yup, my example is incorrect. It switches to polars coordinates on the next line and has the root sign there.

I'm so rusty with this. I'm stuggling to do the division.

EDIT

Its coming back now.