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Let F ( x , y , z ) = ⟨ x − 1 z , y − 1 z , ln ( x y ) ⟩ . Verify that F = ∇ f , where f ( x , y , z ) = z ln ( x y ) . Evaluate ∫ C F ⋅ d r , where r ( t ) = ⟨ e t , e 2 t , t 2 ⟩ for 1 ≤ t ≤ 3 . Evaluate ∫ C F ⋅ d r for any path C from P = ( 1 2 , 4 , 2 ) to Q = ( 2 , 2 , 3 ) contained in the region x > 0 , y > 0 . In part (c), why is it necessary to specify that the path lies in the region where x and y are positive?

User Tguzella
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1 Answer

2 votes

I suppose you mean


\vec F(x,y,z)=\left\langle\frac zx,\frac zy,\ln(xy)\right\rangle

Showing that
f(x,y,z)=z\ln(xy) has gradient equal to
\vec F is trivial.

By virtue of the existence of
f, the gradient theorem applies here, so


\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_((12,4,2))^((2,2,3))\\abla f(x,y,z)\cdot\mathrm d\vec r=f(2,2,3)-f(12,4,2)=3\ln 4-2\ln48=\boxed{-\ln36}

The requirement that
x>0 and
y>0 is a consequence of the domain of the logarithm function. Both have to be positive in order for
\ln(xy) to exist.

User Chfumero
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