Line Integral on a Surface

  • Thread starter Destroxia
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  • #1
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Homework Statement



Use a line integral to find the area of the surface that extends upward from the semicircle ##y=\sqrt{9-x^2}## in the ##xy##-plane to the surface ##z=3x^4y##

Homework Equations



Parametric Equation for Circle:

## x = rcos(t) ##
## y = rsin(t) ##

Line Integral:

## \int_c F \cdot dr ##


The Attempt at a Solution



(Function):
## F = z = 3x^4y ##

(Variables):
## x = 3cos(t) ##
## y = 3sin(t) ##
## dx = -3sin(t)dt ##
## dy = 3cos(t)dt ##

(Intervals):
## -3 ≤ x ≤ 3 ##
## 0 ≤ y ≤ \sqrt{9-x^2} ##
## 0 ≤ t ≤ \pi ## (plugged x-interval into ## x = 3cos(t) ##)

(Solution):
## \int_{0}^{\pi} 3(3cos^{4}(t))(3sin(t))dt = 27\int_{0}^{\pi} cos^4(t)sin(t)dt = * ##
## * = 27 [\frac {-1} {5} cos^5(t) |_{0}^{\pi}] ##
## = 27 [ \frac {2} {5}]##

(Result):
##= \frac {54} {5}##

(Questions):

1. Are there any flaws in this thought process?

2. Since I am switching from substituting x and y with functions of t, do I need some kind of Jacobian in the integral that would take this into account, and affect my final answer?
 

Answers and Replies

  • #2
andrewkirk
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1. Are there any flaws in this thought process?
No, it all looks fine (in principle. I didn't check the integration or the arithmetic).

2. Since I am switching from substituting x and y with functions of t, do I need some kind of Jacobian in the integral that would take this into account, and affect my final answer?
Yes you need a Jacobian, but you already have it in your calcs. The expressions you wrote for ##dx## and ##dy## in terms of ##dt## are representations of the two elements of the Jacobian one gets in transforming from ##x,y## coordinates to ##t## coordinates.
 
  • #3
LCKurtz
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I think you have errors in both your concept and your arithmetic. The first error is that the integral is of the form ##\int_C F \cdot dr##. If that is a dot product it doesn't make sense since ##F## is not a vector. If that ##\cdot## is just a multplication sign then ##dr## would be a scalar, what scalar I wouldn't know.

The trouble is that you should be setting up an integral of the form ##\int_C F(x,y)~ds##. Try again, and be careful about parentheses with your exponents.
 
  • #4
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No, it all looks fine (in principle. I didn't check the integration or the arithmetic).

Yes you need a Jacobian, but you already have it in your calcs. The expressions you wrote for ##dx## and ##dy## in terms of ##dt## are representations of the two elements of the Jacobian one gets in transforming from ##x,y## coordinates to ##t## coordinates.
So, do I have to plug those in anywhere, or do my final calculations already take the Jacobian into account?

I think you have errors in both your concept and your arithmetic. The first error is that the integral is of the form ##\int_C F \cdot dr##. If that is a dot product it doesn't make sense since ##F## is not a vector. If that ##\cdot## is just a multplication sign then ##dr## would be a scalar, what scalar I wouldn't know.
The trouble is that you should be setting up an integral of the form ##\int_C F(x,y)~ds##. Try again, and be careful about parentheses with your exponents.


Okay, my book sets the ##\int_C F(x,y) ds## as ## \int_C F(x,y) ds = \int_C F(x(t),y(t)) \sqrt{(\frac {dx} {dt})^2 + (\frac {dy} {dt})^2} dt ##

So, I set it up as...

## ds = \sqrt{(\frac {dx} {dt})^2 + (\frac {dy} {dt})^2} dt##

## \int_{0}^{\pi} 3(81cos^4(t))(3sin(t)) \sqrt{ (9cos^2(t) + 9sin^2(t))} dt = *##

## * = 2187 \int_{0}^{\pi} cos^4(t)sin(t) dt = 2187 [ \frac {2} {5} ] = \frac {4374} {5} ##

This seems more complete, just wary of my arithmetic a bit. Is this what you were speaking of, about ##\int_C F ds ##?
 
  • #5
LCKurtz
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That's much better, even correct. Might have been a bit easier using polar coordinates using ##ds = 3 d\theta##, which is basically what yours came to.
 

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