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Evaluate ∫C→F⋅d where →F=〈z,3y,0〉 and C is given by →r(t)=〈t,sin(t),cos(t)〉, 0≤t≤π

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To evaluate the line integral ∫C→F⋅d→r, where →F = 〈z, 3y, 0〉 and C is given by →r(t) = 〈t, sin(t), cos(t)〉 for 0 ≤ t ≤ π, we need to compute the dot product →F⋅d→r and integrate it along the curve C.

First, let's find the derivative of →r(t) with respect to t to obtain the tangent vector →T(t):

→r'(t) = 〈1, cos(t), -sin(t)〉

The differential vector d→r is obtained by multiplying →T(t) by dt:

d→r = 〈1, cos(t), -sin(t)〉 dt

Now, let's calculate the dot product →F⋅d→r:

→F⋅d→r = 〈z, 3y, 0〉⋅〈1, cos(t), -sin(t)〉

= z + 3y cos(t)

Substituting the coordinates from →r(t):

→F⋅d→r = t + 3sin(t) cos(t)

Now, we can integrate →F⋅d→r along the curve C. The integral becomes:

∫C→F⋅d = ∫[0,π] (t + 3sin(t) cos(t)) dt

To evaluate this integral, we need to split it into two parts:

∫[0,π] t dt + ∫[0,π] 3sin(t) cos(t) dt

The first integral is straightforward:

∫[0,π] t dt = [t^2/2] evaluated from 0 to π

= (π^2)/2

For the second integral, we can use the trigonometric identity sin(2t) = 2sin(t)cos(t). Then we have:

∫[0,π] 3sin(t) cos(t) dt = (3/2) ∫[0,π] sin(2t) dt

Applying the antiderivative of sin(2t):

= -(3/4) [cos(2t)] evaluated from 0 to π

= -(3/4) (cos(2π) - cos(0))

= -(3/4) (1 - 1)

= 0

Therefore, the line integral evaluates to:

∫C→F⋅d = (π^2)/2 + 0

= (π^2)/2

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