Jaya Legge
2021-10-07
Answered

Evaluate the line integral, where C is the given curve. Integral C x $\mathrm{sin}$ y ds, C is the line segment from (0, 3) to (4, 6)

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bahaistag

Answered 2021-10-08
Author has **101** answers

Recall the way we parametrize the segment between the two points. First, find the vector determined by the given two points:

(4,6)-(0,3)=(4,3)

The starting point is (0,3). The segment can be now parametrized as:

$\gamma \left(t\right)=(0,3)+(4t,3t)=(4t,3t+3),t\in [0,1]$

Next, find the derivative of the parametrization.

${\gamma}^{\prime}\left(t\right)=(4,3)\Rightarrow \left|\left|{\gamma}^{\prime}\left(t\right)\right|\right|=\sqrt{{4}^{2}+{3}^{2}}=5$

Finally, we calculate the integral:

$\int}_{0}^{1}4t\mathrm{sin}(3+3t)\cdot 5dt=20{\int}_{0}^{1}t\mathrm{sin}(3+3t)dt=\frac{20\mathrm{sin}6-60\mathrm{cos}6-20\mathrm{sin}3}{9$

Result:

$\frac{20\mathrm{sin}6-60\mathrm{cos}6-20\mathrm{sin}3}{9}$

(4,6)-(0,3)=(4,3)

The starting point is (0,3). The segment can be now parametrized as:

Next, find the derivative of the parametrization.

Finally, we calculate the integral:

Result:

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