# Line Integral on a Surface

## 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?

andrewkirk
Homework Helper
Gold Member
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.

LCKurtz
Homework Helper
Gold Member
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.

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 ##?

LCKurtz