Iterated Integral Surface Area Problem (with Polor Coordinates)

In summary, the problem involves finding the surface area of the part of a cone lying inside a cylinder. Using the partial derivatives, the surface area is given by an integral over a region defined by polar coordinates. The region can be defined by the equations x^2 - 2x + y^2 = 0 and z = sqrt[(x^2 + y^2)]. The surface area can also be found by using a transformation of the coordinate system. The resulting integral is a constant multiplied by the area of a circle with radius 1, giving a final solution of pi*sqrt(2).
  • #1
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Homework Statement


Find the surface area of the part of the cone z = sqrt[(x^2 + y^2)] lying inside the cylinder x^2 - 2x + y^2 = 0.



2. The attempt at a solution

Partial Derivative x = x/sqrt(x^2 + y^2)
Partial Derivative y = y/sqrt(x^2 + y^2)

so...
sqrt((Partial Derivative Y) ^2 + (Partial Derivative at Y) ^ 2 + 1) =

sqrt[(x^2 + y^2)/(x^2 + y^2) + 1] = sqrt[2]

so...
the surface area =
integral on E of sqrt[2] r dr d(theta),
where E is the region (r, theta)|(0 < r < 2, 0 < theta < pi)

My books solution has the body of the integral as 5, and E as (0<r<1, 0<theta<pi)

I'm pretty sure I've just made an algebraical mistake in the body, but for the limits I'm confused. I don't know how to treat a cylinder that isn't centered at the origin. I thought r was always the distance from the origin... is it really the distance from the center of the cylinder?

Thanks for any help or advice you can give me.
 
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  • #2
alec_tronn said:

Homework Statement


Find the surface area of the part of the cone z = sqrt[(x^2 + y^2)] lying inside the cylinder x^2 - 2x + y^2 = 0.



2. The attempt at a solution

Partial Derivative x = x/sqrt(x^2 + y^2)
Partial Derivative y = y/sqrt(x^2 + y^2)

so...
sqrt((Partial Derivative Y) ^2 + (Partial Derivative at Y) ^ 2 + 1) =

sqrt[(x^2 + y^2)/(x^2 + y^2) + 1] = sqrt[2]

so...
the surface area =
integral on E of sqrt[2] r dr d(theta),
where E is the region (r, theta)|(0 < r < 2, 0 < theta < pi)

My books solution has the body of the integral as 5, and E as (0<r<1, 0<theta<pi)

I'm pretty sure I've just made an algebraical mistake in the body, but for the limits I'm confused. I don't know how to treat a cylinder that isn't centered at the origin. I thought r was always the distance from the origin... is it really the distance from the center of the cylinder?

Thanks for any help or advice you can give me.

That depends on where you put the origin of your coordinate system. x2- 2x+ y2= 0 is the same as (x-1)2+ y2= 1, a circle of radius 1 with center at (1, 0). In "straight" polar coordinates, that circle is given by [itex]r= 2cos(\theta)[/itex]. To integrate over that, let [itex]\theta[/itex] go from 0 to [itex]\pi[/itex] (0 to [itex]2\pi[/itex] would go around the circle twice), r from 0 to [itex]2cos(\theta)[/itex].

Simpler would be to use the transformation [/itex]x= 1+ rcos(/theta)[/itex], [itex]y= r sin(\theta)[/itex], shifting the origin of the coordinate system. Then integrate with [itex]\theta[/itex[/itex] going from 0 to [itex]2\pi[/itex], r from 0 to 1.

Of course, since the integrand is a constant ([itex]\sqrt{2}[/itex], I don't see how your book can have 5), the integral is just that constant times the area of the circle, [itex]\pi\sqrt{2}[/itex].
 
  • #3
alec_tronn said:

Homework Statement


Find the surface area of the part of the cone z = sqrt[(x^2 + y^2)] lying inside the cylinder x^2 - 2x + y^2 = 0.



2. The attempt at a solution

Partial Derivative x = x/sqrt(x^2 + y^2)
Partial Derivative y = y/sqrt(x^2 + y^2)

so...
sqrt((Partial Derivative Y) ^2 + (Partial Derivative at Y) ^ 2 + 1) =

sqrt[(x^2 + y^2)/(x^2 + y^2) + 1] = sqrt[2]

so...
the surface area =
integral on E of sqrt[2] r dr d(theta),
where E is the region (r, theta)|(0 < r < 2, 0 < theta < pi)

My books solution has the body of the integral as 5, and E as (0<r<1, 0<theta<pi)

I'm pretty sure I've just made an algebraical mistake in the body, but for the limits I'm confused. I don't know how to treat a cylinder that isn't centered at the origin. I thought r was always the distance from the origin... is it really the distance from the center of the cylinder?

Thanks for any help or advice you can give me.

That depends on where you put the origin of your coordinate system. x2- 2x+ y2= 0 is the same as
(x-1)2+ y2= 1, a circle of radius 1 with center at (1, 0). In "straight" polar coordinates, that circle is given by [itex]r= 2cos(\theta)[/itex]. To integrate over that, let [itex]\theta[/itex] go from 0 to [itex]\pi[/itex] (0 to [itex]2\pi[/itex] would go around the circle twice), r from 0 to [itex]2cos(\theta)[/itex].

Simpler would be to use the transformation [itex]x= 1+ rcos(/theta)[/itex], [itex]y= r sin(\theta)[/itex], shifting the origin of the coordinate system. Then integrate with [itex]\theta[/itex] going from 0 to [itex]2\pi[/itex], r from 0 to 1.

Of course, since the integrand is a constant ([itex]\sqrt{2}[/itex], I don't see how your book can have 5), the integral is just that constant times the area of the circle, [itex]\pi\sqrt{2}[/itex].
 

1. What is an iterated integral surface area problem with polar coordinates?

An iterated integral surface area problem with polar coordinates is a mathematical problem that involves finding the surface area of a three-dimensional object using polar coordinates. This means that instead of using the traditional x, y, and z coordinates, the object is described using the distance from the origin and the angle from a fixed reference line.

2. How is an iterated integral used to solve this problem?

An iterated integral is used to solve this problem by breaking down the surface area into smaller, simpler parts. The integral is then taken over each part and added together to get the total surface area. When using polar coordinates, the integral is taken with respect to the radius and the angle.

3. What are some common examples of objects that can be solved using an iterated integral surface area problem with polar coordinates?

Some common examples of objects that can be solved using an iterated integral surface area problem with polar coordinates include cones, cylinders, spheres, and other curved surfaces. These objects often have a circular or symmetrical shape that can be described using polar coordinates.

4. What are the benefits of using polar coordinates in an iterated integral surface area problem?

Using polar coordinates in an iterated integral surface area problem can make the calculations simpler and more efficient. This is because the equations for calculating surface area in polar coordinates are often simpler than in Cartesian coordinates, and the symmetry of the objects can be taken advantage of to reduce the number of integrals that need to be evaluated.

5. How does the choice of coordinate system affect the solution to an iterated integral surface area problem?

The choice of coordinate system can greatly affect the solution to an iterated integral surface area problem. Using polar coordinates can make the calculations simpler and more efficient, but it may not always be the best choice. In some cases, Cartesian coordinates may be better suited for the problem at hand, and the choice of coordinate system may also depend on personal preference or the specific requirements of the problem.

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