# Area

• June 5th 2009, 05:02 PM
Apprentice123
Area
A circumference located in the plan $XOY$ have center $C(0,0,0)$ and radius $r > 0$. Determine a parameter for the cylinder formed by all straights that pass by circumference and are parallel to the vector $V(3,0,4)$. Calculate the area of the surface thus obtained, located between plans $\pi 1: z = -4$ and $\pi 2 : z = 4$
• June 6th 2009, 04:15 AM
shawsend
Looks like an elliptic cylinder. Below is a cross-section through the semi-minor axis. The semi-major axis remains at 3 but as you increase the rotation angle, the semi-minor axis becomes less than three and eventually goes to zero in the limit at $\pi/2$. You can calculate the value based on the slope of the vector you specified. We can then take the area of the red cylinder and then just connect the two blue triangular pieces together to form another elliptic cylinder. Then just calculate their areas. Need first to calculate the lengths of the blue and red heights then use the formula for the area of an elliptic cylinder which you can find on mathworld. You can find the coordinates of the green point right? Then you can calculate the lengh (height) of the red line and then the blue line.

• June 7th 2009, 11:19 AM
Apprentice123
Sorry, I do not understand
• June 7th 2009, 11:56 AM
shawsend
Quote:

Originally Posted by Apprentice123
Sorry, I do not understand

. . . dang it. I am way disappointed Apprentice. And It's a really nice plot too. No, not at you but me: We wish to calculate the area of two elliptic cylinders. The red one and the blue one. The red one has height . . . red, and the blue one had height . . . uh, blue. The semi-major axes, $a$ of both are 3, and the semi-minor axis, $b$ is one-half the green line in the new plot below which I've calculated to be $\approx 2.742$. Ok, assume that's all right for now. So we just need to calculate the areas of these two elliptic cylinders. Go now to Mathworld and get the formula for the area:

$A=\int_0^{2\pi}\int_0^{h} \sqrt{a^2\sin^2\theta+b^2\cos^2 \theta} dh d\theta$

I'm not entirely sure of this but let's assume that's correct for now. So I calculated the red length is $32/5$. So the area of the red cylinder is:

$A=\int_0^{2\pi}\int_0^{32/5} \sqrt{a^2\sin^2\theta+b^2\cos^2 \theta} dh d\theta\approx 115.5$

which you can either calculate numerically like I did, or express the integral in terms of a complete elliptic integral of the second kind:

$\text{E}(k)=\int_0^{\pi/2}\sqrt{1-k\sin^2(\theta)}d\theta$

Now, suppose it was just a cylinder of height $32/5$ and radius 3. Then the area would be $6\pi 32/5\approx 120$. Hey, that's close. If it were 1000 or so then I'd say who in here knows this better than me?. But that's close and gives me confidence I'm . . .uh, close. Can you do the blue one?
• June 7th 2009, 02:02 PM
Apprentice123
Quote:

Originally Posted by shawsend
. . . dang it. I am way disappointed Apprentice. And It's a really nice plot too. No, not at you but me: We wish to calculate the area of two elliptic cylinders. The red one and the blue one. The red one has height . . . red, and the blue one had height . . . uh, blue. The semi-major axes, $a$ of both are 3, and the semi-minor axis, $b$ is the green line in the new plot below which I've calculated to be $\approx 2.742$. Ok, assume that's all right for now. So we just need to calculate the areas of these two elliptic cylinders. Go now to Mathworld and get the formula for the area:

$A=\int_0^{2\pi}\int_0^{h} \sqrt{a^2\sin^2\theta+b^2\cos^2 \theta} dh d\theta$

I'm not entirely sure of this but let's assume that's correct for now. So I calculated the red length is $32/5$. So the area of the red cylinder is:

$A=\int_0^{2\pi}\int_0^{32/5} \sqrt{a^2\sin^2\theta+b^2\cos^2 \theta} dh d\theta\approx 115.5$

which you can either calculate numerically like I did, or express the integral in terms of a complete elliptic integral of the second kind:

$\text{E}(k)=\int_0^{\pi/2}\sqrt{1-k\sin^2(\theta)}d\theta$

Now, suppose it was just a cylinder of height $32/5$ and radius 3. Then the area would be $6\pi 32/5\approx 120$. Hey, that's close. If it were 1000 or so then I'd say who in here knows this better than me?. But that's close and gives me confidence I'm . . .uh, close. Can you do the blue one?

Ok. Thank you