Line Integral on a Surface

[Destroxia]

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]

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]

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.

 

[Destroxia]

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]

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.