Can anyone confirm this possible error in 'Vector Calculus' by Matthews?

In summary, the solution for the surface integral of u dot n over S is (1,2y,-1)dxdy, and this is due to the fact that the surface is parameterized as ##\vec R(x,y) = \langle x, y, x+y^2\rangle##.
  • #1
Syrus
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Homework Statement



Find the surface integral of u dot n over S where S is the part of the surface z = x + y2 with z<0 and x>-1, u is the vector field u = (2y+x,-1,0) and n has a negative z component.

Homework Equations



In the text leading up to the end-of-chapter exercises (where this question arises), n is specified as a unit normal vector.

The Attempt at a Solution



The solution claims that ndS = (1,2y,-1)dxdy. But isn't this ignoring the fact that n should be a unit vector? Obviously, the position vector for the surface in parameter form is r = (x,y,x+y2), and n is obtained by taking the partial derivative of r with respect to y, the partial derivative of r with respect to x, and forming the cross product of the two (in that order). But then, shouldn't this resulting vector be divided by the magnitude of the cross product- which, by my calculations happens to be √(2 + 4y2), which is never 1?
 
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  • #2
Syrus said:

Homework Statement



Find the surface integral of u dot n over S where S is the part of the surface z = x + y2 with z<0 and x>-1, u is the vector field u = (2y+x,-1,0) and n has a negative z component.



Homework Equations



In the text leading up to the end-of-chapter exercises (where this question arises), n is specified as a unit normal vector.

The Attempt at a Solution



The solution claims that ndS = (1,2y,-1)dxdy. But isn't this ignoring the fact that n should be a unit vector? Obviously, the position vector for the surface in parameter form is r = (x,y,x+y2), and n is obtained by taking the partial derivative of r with respect to y, the partial derivative of r with respect to x, and forming the cross product of the two (in that order). But then, shouldn't this resulting vector be divided by the magnitude of the cross product- which, by my calculations happens to be √(2 + 4y2), which is never 1?

Your surface is parameterized as ##\vec R(x,y) = \langle x, y, x+y^2\rangle##. Read my post #13 in this thread:

https://www.physicsforums.com/showthread.php?t=611873

with u = x and v = y and you will see why the book is correct.
 
  • #3
Understood. Thank you, LCKurtz.
 

FAQ: Can anyone confirm this possible error in 'Vector Calculus' by Matthews?

What is the possible error in 'Vector Calculus' by Matthews?

The possible error in 'Vector Calculus' by Matthews is a mistake in the equation for calculating the divergence of a vector field in spherical coordinates. It is believed that the author may have mistakenly used the wrong coordinate system, resulting in an incorrect formula.

How was this possible error discovered?

This possible error was discovered by a group of students who were studying the book for their vector calculus course. They noticed that the equation for divergence in spherical coordinates did not match with what they were learning in class, prompting them to investigate further.

Has the author been notified about this possible error?

Yes, the author has been notified about this possible error. The students who discovered the mistake have reached out to the author through email and social media, providing evidence and explanation for the error.

Are there any other errors in 'Vector Calculus' by Matthews?

At this time, there is no evidence of any other errors in the book. However, it is always important to carefully review and double-check mathematical equations and formulas to ensure accuracy.

Will there be a correction or update to the book?

It is up to the author and the publisher to decide if a correction or update will be made to the book. They may choose to release a new edition or make a correction in future printings. In the meantime, it is recommended to be cautious and double-check any equations or formulas from the book.

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