Vector Calculus, setting up surface area integral.

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Homework Help Overview

The problem involves finding the surface area of a portion of a cone defined by the equation z^2 = x^2 + y^2, constrained within a cylinder described by y^2 + z^2 ≤ 49. The original poster expresses difficulty in setting up the limits of integration for the surface area integral.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the intersection of the cone and cylinder, with one suggesting the use of parametric equations and another proposing a change to polar coordinates for integration. There is also a mention of the elliptical shape of the intersection when viewed from above.

Discussion Status

Some participants have offered guidance on using polar coordinates and have noted the relationship between the area of the ellipse and the integral setup. There appears to be a productive exploration of different approaches, though no explicit consensus is reached on the method.

Contextual Notes

Participants mention constraints such as the problem not being on any tests and the original poster's uncertainty about the limits of integration. There is also a reference to the area of an ellipse, which may influence the approach taken.

jonathanm111
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The question goes like: find the SA of the portion S of the cone z^2 =x^2 +y^2 where z>=0 contained within the cylinder y^2+z^2<=49
attempt.jpg

this is my attempt using the formula for SA, I could switch to parametric eqns, but even then I'd have hard time setting up limits of integration.
 

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I haven't worked it all out, but here's what I would try. The two surfaces intersect where their ##z## values are equal. So substituting ##z^2## on the cone for ##z^2## on the cylinder gives ##x^2+2y^2 = 49##, or$$
\frac{x^2}{49} + \frac{y^2}{\frac{49}{2}} = 1$$This tells you if you look straight down on the surfaces, their intersection curve looks like an ellipse and, in fact, the ##xy## domain is the interior of that ellipse. So you can set it all up in the ##x## and ##y## variables and integrate over the interior of that ellipse. If you are lucky it will come out real simple, otherwise, if you have to work the integral out, you might want to change coordinates to polar like coordinates ##x = 7\cos\theta,~ y = \frac 7 {\sqrt 2} \sin\theta##. Here's a sketch:
graphs.jpg
 

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I got it
solution.jpg


you were right, polar coordinates were the way to go. this was a strange problem, professor said it wouldn't be on any tests anyways.
Thank you.
 

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Yes, it looks like you have it. But from your previous work you have ##dS=\sqrt 2~dydx## so your integral is##\iint_{ellipse}~\sqrt 2~dydx = \sqrt 2 \text{Area}## If you know the area of an ellipse with semi-axes ##a## and ##b## is ##\pi a b## you have from the equation of the ellipse ##\sqrt 2 \cdot 7 \cdot \frac 7 {\sqrt 2}=49\pi##, saving a bit of work.
 
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LCKurtz said:
Yes, it looks like you have it. But from your previous work you have ##dS=\sqrt 2~dydx## so your integral is##\iint_{ellipse}~\sqrt 2~dydx = \sqrt 2 \text{Area}## If you know the area of an ellipse with semi=axes ##a## and ##b## is ##\pi a b## you have from the equation of the ellipse ##\sqrt 2 \cdot 7 \cdot \frac 7 {\sqrt 2}=49\pi##, saving a bit of work.
I believe that's how they wanted us to solve it haha. nice find
 

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