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Evaluate the line integral along the path \(C\) given by \(x=2 t, y=10 t,\) where \(0 \leq t \leq 1\) \(\int_{C}(3 y-x) d x+y^{2} d y\)

Short Answer

Expert verified
The line integral over the path \(C\) is 20

Step by step solution

01

Parameterize the Vector Field

Rewrite the integral in terms of the given parameter \(t\). Replace \(x\) and \(y\) in the integral with the given equations \(x=2t\) and \(y=10t\). So the integral \(\int_{C}(3y - x)dx + y^2dy\) becomes \(\int_{C}(3*10t - 2t)d(2t) + (10t)^2d(10t)\)
02

Simplify the Integral

Now the integral can be simplified to \(\int_{0}^{1}(60t - 4t^2)dt + \int_{0}^{1}100t^2dt\). Here, the limits of the integral are set to 0 and 1 after identifying the limits of \(t\) in the question
03

Evaluate the two separate integrals

Evaluate the two separate integrals one by one. For the first integral, using the power rule for antiderivatives, \(\int_{0}^{1}(60t - 4t^2)dt\) evaluates to \([30t^2 - \frac{4}{3}t^3]_{0}^{1}\). Similarly, the other integral evaluates to \(\int_{0}^{1}100t^2dt = [\frac{100}{3}t^3]_{0}^{1}\)
04

Substitute the limits

Now plug in the limits i.e., 1 and 0 into the evaluated integrals. The final answer will be the summation of those two evaluated integrals
05

Do the actual computations

Doing the actual computations should now give you the answer. After evaluating all of this, you will have \(30(1)^2 - \frac{4}{3}(1)^3 + \frac{100}{3}(1)^3 - 0\) which simplifies to \(20\)

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Most popular questions from this chapter

Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) for each curve. Discuss the orientation of the curve and its effect on the value of the integral. \(\mathbf{F}(x, y)=x^{2} y \mathbf{i}+x y^{3 / 2} \mathbf{j}\) (a) \(\mathbf{r}_{1}(t)=(t+1) \mathbf{i}+t^{2} \mathbf{j}, \quad 0 \leq t \leq 2\) (b) \(\mathbf{r}_{2}(t)=(1+2 \cos t) \mathbf{i}+\left(4 \cos ^{2} t\right) \mathbf{j}, \quad 0 \leq t \leq \pi / 2\)

In Exercises \(5-8,\) evaluate the line integral along the given path. \(\int_{C}(x-y) d s\) \(C: \mathbf{r}(t)=4 t \mathbf{i}+3 t \mathbf{j}\) \(0 \leq t \leq 2\)

Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) where \(C\) is represented by \(\mathbf{r}(t)\) \(\mathbf{F}(x, y)=3 x \mathbf{i}+4 y \mathbf{j}\) \(\quad C: \mathbf{r}(t)=2 \cos t \mathbf{i}+2 \sin t \mathbf{j}, \quad 0 \leq t \leq \pi / 2\)

Determine the value of \(c\) such that the work done by the force field \(\mathbf{F}(x, y)=15\left[\left(4-x^{2} y\right) \mathbf{i}-x y \mathbf{j}\right]\) on an object moving along the parabolic path \(y=c\left(1-x^{2}\right)\) between the points (-1,0) and (1,0) is a minimum. Compare the result with the work required to move the object along the straight-line path connecting the points.

In Exercises \(43-46,\) demonstrate the property that \(\int_{C} \mathbf{F} \cdot d \mathbf{r}=\mathbf{0}\) regardless of the initial and terminal points of \(C,\) if the tangent vector \(\mathbf{r}^{\prime}(t)\) is orthogonal to the force field \(\mathbf{F}\) \(\mathbf{F}(x, y)=y \mathbf{i}-x \mathbf{j}\) \(C: \mathbf{r}(t)=t \mathbf{i}-2 t \mathbf{j}\)

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