Line Integral on Sphere

Calculus Level 3

Consider the following curve, where the coordinates of each point on the curve are ( x , y , z ) (x,y,z) :

x = cos θ sin ϕ y = sin θ sin ϕ z = cos ϕ θ = α ϕ = π 2 α 0 α π 2 x = \cos \theta \, \sin \phi \\ y = \sin \theta \, \sin \phi \\ z = \cos \phi \\ \theta = \alpha \\ \phi = \frac{\pi}{2} - \alpha \\ 0 \leq \alpha \leq \frac{\pi}{2}

There is also a vector field present at all points in space:

F = ( F x , F y , F z ) = ( x , y 2 , z 3 ) \vec{F} = (F_x,F_y,F_z) = (x,y^2,z^3)

What is the absolute value of the line integral of the vector field over the curve?

C F d = ? \Big | \int_C \vec{F} \cdot \vec{d \ell} \, \Big | = ?


The answer is 0.25.

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1 solution

Otto Bretscher
Jan 17, 2019

Greetings from Grenada, Steven!

The path runs from A ( 1 , 0 , 0 ) A(1,0,0) to B ( 0 , 0 , 1 ) B(0,0,1) , and F \vec{F} has a potential f ( x , y , z ) = x 2 2 + y 3 3 + z 4 4 f(x,y,z)=\frac{x^2}{2}+\frac{y^3}{3}+\frac{z^4}{4} so the line integral is f ( B ) f ( A ) = 1 4 f(B)-f(A)=-\frac{1}{4} and the answer is 0.25 \boxed{0.25} .

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