Integrating zdS on a Sphere Centered at Origin

[Smusko]

I will denote vectors in bold.

Homework Statement

Show that the curve C given by

r=a*Cos(t)Sin(t)i+a*Sin²(t)j+a*Cos(t)k ( 0=<t=<pi/2 )
lies on a sphere centred at the origin.

Find $$\int zdS$$ under C

*edit* There is a huge gap here and the equation has dissapered for me. But what it said was Integral(zdS) under C

Homework Equations

The solution is (a²/2)*(sqrt(2)+ln(1+sqrt(2))

The Attempt at a Solution

To show that the curve was on a sphere centred at the origin I just put in various values for t onto the curve and plotted it onto a graph.
Maybe there's a way to do it algebraically. But that is not the part I'm looking for help with.

I want help figuring out what to do with the Z in the integral and calculating it.

I got dS = |dr / dt|*dt = a*cos(t)dt

I am totally lost how to parametrize Z or figuring out the domain. The book I use gives no clear example on what to do, or it is just wooshing over my head.

I have tried grasping straws but nothing comes close to the answer given by the solution section.

Help is much appreciated. :)

*edit*

Latex seems to be totally bugging out on me. Weird suff shows up when I refresh, Stuff I have edited away keeps coming back, and now there is a huge gap in the post. Tell me if you see the same thing or if its just me.

 

[Metaleer]

To show that the given curve lies on a sphere centered at the origin, all you need to do is calculate the distance between the origin and any point that this curve defines. You should see that the result is not a function of the parameter.

If you mean the scalar line integral of

$$\int_\mathcal{C} z ds,$$​

you have to remember that the differential element $$ds$$ is

$$ds = \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}dt$$​

where your curve $$\mathcal{C}$$ is given by

$$\mathbf{r} = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k}$$​

and from your curve, $$z = a \cos t$$, so all you need to do is plug in these values in the integral and integrate from $$0$$ to $$\pi/2$$.

Hope this helps.

 

[Smusko]

Yeah, helps allot. I have done it that way and failed but I have a knack for making stupid tiny mistakes, like forgetting a minus sign or something.
So Il try again. It is nice to know that this stuff is not completely beyond me. :)

 

[Metaleer]

Yeah, helps allot. I have done it that way and failed but I have a knack for making stupid tiny mistakes, like forgetting a minus sign or something.

As a professor of mine says, it happens in the best families. :)

So Il try again. It is nice to know that this stuff is not completely beyond me. :)

The method should work. There's also a tiny possibility that the book's answer is wrong, which is why after trying to solve a problem a few times with no luck, I always tend to resort to WolframAlpha to see if I bungled the integral somewhere, or if indeed the book's wrong.

 

[Smusko]

I have checked with Wolfram on all the derivatives and integrals. The dS must be equal to a*cos(t)dt
and if it is like you say that you put z=a*cos(t) and integrate from 0 to pi/2 then the integral should look like this:
a²*Integral(cos²(t))dt = a²((t/2) + (1/4)*sin(2t))

Evaluated from 0 to pi/2 you get a²*pi/4

Right or wrong i know how to calculate them :) thank you.

 

[Metaleer]

The reason you're not getting the book's answer is because you botched $$ds$$. Using the formula I provided, I get

$$ds = a \sqrt{\sin^2 t + 1},$$​

assuming that $$a$$ is a positive constant. Using this value and the corresponding value of $$z$$, the book's answer follows.

 

[Smusko]

Dang, thought. Missed the ********* **** **** minus sign. I got sqrt(1-Sin²(t)) because (dz/dt) = -aSin(t).
Thanks.