Iterated Integral Surface Area Problem (with Polor Coordinates)

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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.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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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.
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
HallsofIvy
Science Advisor
Homework Helper
41,833
961

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].
 

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