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Chapter 15: Problem 21

Evaluate the line integral along the curve C. $$ \begin{array}{l}{\int_{C}-y d x+x d y} \\ {C: y^{2}=3 x \text { from }(3,3) \text { to }(0,0)}\end{array} $$

Short Answer

Expert verified
The value of the line integral is 3.

Step by step solution

01

Parametrize the Curve

The given curve is \( y^2 = 3x \). We can parametrize it in terms of \( y \) by solving for \( x \). Thus, \( x = \frac{y^2}{3} \). Let \( y = t \) where \( t \) ranges from \( 3 \) to \( 0 \). Then \( x = \frac{t^2}{3} \) and our parametrization becomes \((x(t), y(t)) = \left(\frac{t^2}{3}, t\right)\).
02

Express the Differential Terms

With the parametrization \( x(t) = \frac{t^2}{3} \) and \( y(t) = t \), calculate the differentials \( dx \) and \( dy \). Then, \( dx = \frac{d}{dt}\left(\frac{t^2}{3}\right) dt = \frac{2t}{3} dt \) and \( dy = \frac{d}{dt}(t) dt = dt \).
03

Substitute in the Line Integral

Substitute the parametrized forms and differentials into the line integral: \( \int_C -y dx + x dy = \int_{3}^{0} -t \cdot \frac{2t}{3} dt + \frac{t^2}{3} \cdot dt \). Simplify this to: \( \int_{3}^{0} \left( -\frac{2t^2}{3} + \frac{t^2}{3} \right) dt \). Further simplify to \( \int_{3}^{0} -\frac{t^2}{3} dt \).
04

Evaluate the Integral

Evaluate the integral: \( \int_{3}^{0} -\frac{t^2}{3} dt = -\frac{1}{3} \int_{3}^{0} t^2 dt \). Find the antiderivative of \( t^2 \), which is \( \frac{t^3}{3} \), and then compute: \( -\frac{1}{3} \left[ \frac{t^3}{3} \right]_{3}^{0} = -\frac{1}{3} \left( \frac{0^3}{3} - \frac{3^3}{3} \right) \).
05

Simplify the Result

Simplify the computation in the previous step: \( -\frac{1}{3} \left( 0 - 9 \right) = -\frac{1}{3} \times (-9) = 3 \). The value of the line integral is thus 3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Curve Parametrization
Beginning with the concept of **curve parametrization**, it is crucial to express a curve in terms of a single parameter, often labeled as 't'.
For the curve given by \( y^2 = 3x \), the process involves solving for one variable in terms of the other. We choose \( y = t \), which simplifies the equation to \( x = \frac{t^2}{3} \).
Thus, the parametrized form of the curve is given by the set of functions \( (x(t), y(t)) = \left( \frac{t^2}{3}, t \right) \).

This makes it easier to evaluate integrals as it transforms the curve into a function of 't'. Here, 't' ranges from 3 to 0 as we move from the starting point \((3,3)\) to the ending point \((0,0)\).
This parametrization simplifies the integration process tackling the complexity of the curve by using the parameter 't' as a bridge.
Differential Calculus
In the context of line integrals, differential calculus plays a significant role. Once the curve is parametrized, the next primary task is to compute the differentials \( dx \) and \( dy \).
With functions \( x(t) = \frac{t^2}{3} \) and \( y(t) = t \), the differentials are obtained by differentiating these with respect to 't'.
  • To find \( dx \), differentiate \( x(t) \) to get \( dx = \frac{2t}{3} dt \).
  • Similarly, for \( dy \), we differentiate \( y(t) \) which results in \( dy = dt \).
These differentials are vital as they allow us to express the integrand in terms of a single variable, thereby simplifying the integral significantly.
Integral Evaluation
After parametrizing the curve and computing the differentials, the process of **integral evaluation** focuses on substituting these into the line integral. We start with the integral \( \int_{C} -y \, dx + x \, dy \) and substitute our expressions from parametrization.
This transforms the problem into an integral in terms of 't', \( \int_{3}^{0} \left( -\frac{2t^2}{3} + \frac{t^2}{3} \right) dt \), and simplifies further to \( \int_{3}^{0} -\frac{t^2}{3} dt \).
Next, we evaluate the integral by finding the antiderivative of \( t^2 \), which is \( \frac{t^3}{3} \).
The evaluation involves applying the limits '3' and '0', resulting in:
  • \( -\frac{1}{3} \left[ \frac{t^3}{3} \right]_{3}^{0} \)
  • This simplifies to \( -\frac{1}{3} \left( 0 - 9 \right) = -\frac{1}{3} \times (-9) = 3 \).
Thus, careful integration leads us to the final result of 3 for the line integral.

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Most popular questions from this chapter

Use a CAS to approximate the mass of the curved lamina \(z=e^{-x^{2}-y^{2}}\) that lies above the region in the \(x y\) -plane enclosed by \(x^{2}+y^{2}=9\) given that the density function is \(\delta(x, y, z)=\sqrt{x^{2}+y^{2}}\).

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Evaluate the surface integral $$ \iint_{\sigma} f(x, y, z) d S $$ \(f(x, y, z)=\left(x^{2}+y^{2}\right) z ; \sigma\) is the portion of the sphere \(x^{2}+y^{2}+z^{2}=4\) above the plane \(z=1\)
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