# 2 cylinders intersect, area of resulting parametric surface

• kostoglotov
In summary, the conversation discusses a problem involving a surface and a parametric vector equation. The individual starts by dividing the surface into smaller sections to simplify the problem. They use symmetry to further reduce the surface to one sixteenth of the original. The resulting equation is then integrated to find the surface area, which turns out to be 16.
kostoglotov

## Homework Statement

I want to know if I got the answer correct and if my reasoning is sound. The text answers and solutions manual only gives answers/solutions for odd numbered problems.

Here is the problem:

And a direct link to the imgur page: http://i.imgur.com/Tko1xFh.png

## Homework Equations

$$A(S) = \int \int_D \left|\vec{r}_u\times\vec{r}_v\right|dA$$

## The Attempt at a Solution

Instead of trying to come up with a parametric vector equation for the entire surface, I started subdividing the surface as much as possible using symmetry. I got it down to one sixteenth of the whole surface. Basically, just consider the surface in the first octant, and then divide that in two again along the line y = x.

By doing that I get the following:

$$A(S) = 16 \times \int\int_D\left|\vec{r}_x\times\vec{r}_y\right|dA$$

where $\vec{r} = \left< x,y,\sqrt{1-x^2}\right>$ and $D = \{(x,y)|0\leq x \leq 1, 0 \leq y \leq x\}$

so

$\vec{r}_x = \langle 1,0,\frac{-x}{\sqrt{1-x^2}} \rangle$ and $\vec{r}_y = \langle 0,1,0 \rangle$

$$\vec{r}_x \times \vec{r}_y = \langle \frac{x}{\sqrt{1-x^2}},0,1 \rangle$$

$$\left| \vec{r}_x \times \vec{r}_y \right| = \sqrt{1+\frac{x^2}{1-x^2}} = \frac{1}{1-x^2}$$

Now

$$A(S) = 16 \times \int_0^1\int_0^x \frac{1}{1-x^2} dydx$$

$$A(S) = 16 \times \int_0^1 \frac{x}{1-x^2} dx = -16 \times\left[\sqrt{1-x^2}\right]_0^1 = 16$$

kostoglotov said:

## Homework Statement

I want to know if I got the answer correct and if my reasoning is sound. The text answers and solutions manual only gives answers/solutions for odd numbered problems.

Here is the problem:

And a direct link to the imgur page: http://i.imgur.com/Tko1xFh.png

## Homework Equations

$$A(S) = \int \int_D \left|\vec{r}_u\times\vec{r}_v\right|dA$$

## The Attempt at a Solution

Instead of trying to come up with a parametric vector equation for the entire surface, I started subdividing the surface as much as possible using symmetry. I got it down to one sixteenth of the whole surface. Basically, just consider the surface in the first octant, and then divide that in two again along the line y = x.

By doing that I get the following:

$$A(S) = 16 \times \int\int_D\left|\vec{r}_x\times\vec{r}_y\right|dA$$

where $\vec{r} = \left< x,y,\sqrt{1-x^2}\right>$ and $D = \{(x,y)|0\leq x \leq 1, 0 \leq y \leq x\}$

so

$\vec{r}_x = \langle 1,0,\frac{-x}{\sqrt{1-x^2}} \rangle$ and $\vec{r}_y = \langle 0,1,0 \rangle$

$$\vec{r}_x \times \vec{r}_y = \langle \frac{x}{\sqrt{1-x^2}},0,1 \rangle$$

$$\left| \vec{r}_x \times \vec{r}_y \right| = \sqrt{1+\frac{x^2}{1-x^2}} = \frac{1}{1-x^2}$$
Shouldn't that be$$\frac{1}{\sqrt{1-x^2}}$$

kostoglotov
LCKurtz said:
Shouldn't that be$$\frac{1}{\sqrt{1-x^2}}$$

Typo :) In my original work that typo isn't there.

I haven't tried to follow your working (yet), but a mental calculation using a simpler method gives me 16 also.

kostoglotov
haruspex said:
I haven't tried to follow your working (yet), but a mental calculation using a simpler method gives me 16 also.
My method:
Convert the xz plane to polar, measuring theta from lowest point.
Consider face on the positive x side.
Between ##\theta## and ##\theta+d\theta##, there is a rectangle measuring ##r.d\theta## by ##2r\sin(\theta)##. Integrate 0 to ##\pi## and multiply by 4.

## 1. What is a parametric surface?

A parametric surface is a mathematical concept that describes a surface in three-dimensional space as a set of points that can be represented by a set of parameters. These parameters can be defined by equations or functions, and they allow for a more flexible and general representation of surfaces compared to traditional methods.

## 2. How do you find the area of a parametric surface?

The area of a parametric surface can be found by using a surface integral. This involves breaking the surface into small, infinitesimal pieces and adding up their areas using a double integral. The integral can be solved by using the parametric equations that define the surface.

## 3. What is the significance of two cylinders intersecting in a parametric surface?

The intersection of two cylinders in a parametric surface is significant because it creates a unique shape that can be described by equations or functions. This shape can have interesting properties and applications in mathematics, physics, and engineering.

## 4. How does the area of the resulting parametric surface change as the cylinders intersect at different angles?

The area of the resulting parametric surface will vary depending on the angle at which the cylinders intersect. If the cylinders intersect at a sharp angle, the resulting surface will have a larger area compared to when they intersect at a more gradual angle. This can be seen by visualizing the intersection of two cones, which are essentially just tilted cylinders.

## 5. Can the area of a resulting parametric surface be negative?

No, the area of a resulting parametric surface cannot be negative. The concept of area is always positive, and even if the surface has a concave shape, the area can be calculated by using the magnitude of the surface integral. Therefore, the area of a resulting parametric surface will always be a positive value.

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