Question

I'm trying to find the area of the portion of a tilted plane inside a cylinder using parametrization, posted below momentarily.

Answer 1

http://i.imgur.com/7nHAfPy.png

How exactly should I set up the vector r? I'm guessing I should base the parametrization of r off of the cylinder?

Answer 2

\[R(u,v) = \ ?\]

Answer 3

@FibonacciChick666 , could you help me out on this? Just trying to figure out the parametrization.

Answer 4

I'm working on a paper, but here is a really good explanation someone else wrote http://math.stackexchange.com/questions/817869/surface-integral-the-area-of-a-plane-inside-a-cylinder

Answer 5

ty!

Answer 6

np, good luck on the assignment

Answer 7

I'm trying to take a shot at this using a calculus-based approach, and have come up with this. @Zarkon , could you help me out with this/does this setup look okay for the parametrization of the curve intersecting the cylinder and the tilted plane?

http://i.imgur.com/xhsfoyR.png

Answer 8

\[R(r,\theta)=r \cos(\theta)i + r \sin(\theta)j + (1-\frac{1}{2}r \sin(\theta))k.\]

Answer 9

\[R_r=\cos(\theta)+\sin(\theta)j-\frac{1}{2}\sin(\theta)k.\]

Answer 10

\[R_\theta = -r \sin(\theta)i+r \cos(\theta)j-\frac{1}{2}r \cos(\theta)k\]

Answer 11

@wio , could you help me out with this when you get a chance? Thanks so far.

Answer 12

This is a tricky one, no doubt.

Answer 13

If you use cylindrical coords then...

Answer 14

Sorry, was working on another problem.

Wait, why isn't it 2z = 2 - y?

Answer 15

\[
y+2z=2\implies z=1-\frac y2 = 1-\frac{r\sin\theta}{2}
\]

Answer 16

\[y+2z=2; \ \ \ 2z = 2-y; \ \ \ z = \frac{2-y}{2}\]

Answer 17

Either way is fine.

Answer 18

Your partial derivatives are fine as well.

Answer 19

Alright, cool, taking a shot at computing the cross product now between the two vectors that come from it. How do I set up the, uh....the determinant matrix in laTeX in this kinda environment? I think I'll just write it out

Answer 20

\[
\left|\mathbf \Phi_{r}\times \mathbf \Phi_{\theta}\right|=
\begin{vmatrix}
\mathbf i &\mathbf j &\mathbf k \\
\cos(\theta)&\sin(\theta)&-\frac 12 \sin(\theta)\\
-r\sin(\theta)&r\cos(\theta)&-\frac 12r \cos(\theta)
\end{vmatrix}
\]

Answer 21

Whoop, here we go: http://i.imgur.com/rLXc9Mk.png

Answer 22

(Just double checking the vector before taking the magnitude o fit)

Answer 23

Alright, now, taking the magnitude looks super ugly....takign a shot at it now.

Answer 24

Now that I think about it, I wonder if the normal vectors on the plane would be different.

Answer 25

Hm?

Answer 26

The normal vectors on the plane should be the same direction as the normal vectors on the disk we parametrized.

Answer 27

It's just maybe the magnitude is different.

Answer 28

I dunno, I'm being clueless atm and don't catch your drift; should I be taking the magnitude of that cross product as it stands?

Answer 29

Just keep doing what you were doing.

Answer 30

http://i.imgur.com/IVXMubo.png

Answer 31

Not putting that all under a square root and summing it,

Answer 32

Okay, how the hell are you getting that?

Answer 33

Lul

Answer 34

Do we agree that the vector, before taking its magnitude, is http://i.imgur.com/BTgWUiw.png

Answer 35

\[
\begin{split}
\left|\mathbf \Phi_{r}\times \mathbf \Phi_{\theta}\right|
&=&
\begin{vmatrix}
\mathbf i &\mathbf j &\mathbf k \\
\cos(\theta)&\sin(\theta)&-\frac 12 \sin(\theta)\\
-r\sin(\theta)&r\cos(\theta)&-\frac 12r \cos(\theta)
\end{vmatrix} \\
&=&+ \mathbf i\left(-\frac 12\sin(\theta) r\cos(\theta)+\frac 12\sin(\theta) r\cos(\theta)\right)\\
&&-\mathbf j\left(-\frac 12r\cos^2(\theta)-\frac 12r\sin^2(\theta)\right)\\
&&+\mathbf k\left(r\cos^2(\theta)+r\sin^2(\theta)\right) \\
&=&\left|\frac 12 r\mathbf j+r\mathbf k\right| = \sqrt{\frac 14r^2+r^2} = \frac{\sqrt 5}{2}r
\end{split}
\]

Answer 36

J component is one minus the other though, right? Minus a minus?

Answer 37

(but yeah I get your point lol)

Answer 38

The term is positive, but it gets subtracted.

Answer 39

Wait, the term is *negative*, but it gets subtracted, right, so it makes it a positive. One sec.

Answer 40

OH, I missed a minus sign in my original rprlbem neverMKFEWOHOJA

Answer 41

I gotchu

Answer 42

I'm bad at maths without paper, lol.

Answer 43

Alright, well, cool, got the general setup. Now going to go work some other problems, just need to watch for algebra errors. Thank you.