Vector Calculus, setting up surface area integral.

In summary, the question asks if the surface area of the portion of the cone z^2 =x^2 +y^2 where z>=0 contained within the cylinder y^2+z^2<=49 is equal to 49. If you use polar coordinates, you can find the area by integrating over the interior of the ellipse.
  • #1
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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  • #2
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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  • #3
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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  • #4
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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  • #5
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
 

What is vector calculus?

Vector calculus is a branch of mathematics that deals with vector fields, which are functions that assign a vector to every point in a given space, and the operations that can be performed on these vector fields.

What is a surface area integral?

A surface area integral is a type of integral in vector calculus that is used to calculate the surface area of a three-dimensional object. It involves integrating a scalar or vector field over a given surface.

How do you set up a surface area integral?

To set up a surface area integral, you first need to determine the limits of integration, which define the boundaries of the surface. Next, you need to determine the integrand, which is the function that is being integrated. Finally, you need to choose the appropriate surface element, which is a small piece of the surface that is being integrated over.

What are some applications of surface area integrals?

Surface area integrals are used in various fields, such as physics, engineering, and computer graphics, to calculate the surface area of objects, determine flux through a surface, and calculate work done by a force on a surface, among other applications.

What is the difference between a surface area integral and a line integral?

The main difference between a surface area integral and a line integral is the dimension of the object being integrated over. A surface area integral is used to integrate over a two-dimensional surface, while a line integral is used to integrate over a one-dimensional curve.

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