Problem 26
Question
Evaluate \(\int_{C} y d x+z d y+x d z\) on the given curve \(C\) between \((0,0,0)\) and \((6,8,5)\). $$ x=3 t, y=t^{3}, z=\frac{5}{4} t^{2}, \quad 0 \leq t \leq 2 $$
Step-by-Step Solution
Verified Answer
The value of the integral is 44.
1Step 1: Parametrize the Curve
Given the parametrization of the curve \( x = 3t \), \( y = t^3 \), and \( z = \frac{5}{4}t^2 \), determine the differential forms. Here \( dx = 3 \cdot dt \), \( dy = 3t^2 \cdot dt \), and \( dz = \frac{5}{2}t \cdot dt \).
2Step 2: Setup the Integral
Substitute the differentials and the parametric equations into the integral. We have:\[\int_{C} y \, dx + z \, dy + x \, dz = \int_{0}^{2} (t^3 \cdot 3) + \left(\frac{5}{4}t^2 \cdot 3t^2\right) + (3t \cdot \frac{5}{2}t) \,dt.\]
3Step 3: Simplify the Expression
Simplify the integrand:\[3t^3 + \frac{15}{4}t^4 + \frac{15}{2}t^2 \cdot dt.\]
4Step 4: Integrate Term by Term
Perform the integration for each term separately:- The integral for \(3t^3\) is \(\frac{3}{4}t^4\).- The integral for \(\frac{15}{4}t^4\) is \(\frac{15}{20}t^5\).- The integral for \(\frac{15}{2}t^2\) is \(\frac{15}{6}t^3\).
5Step 5: Evaluate the Definite Integral
Substitute the limits of integration \(0\) and \(2\) into the antiderivatives:- For \(\frac{3}{4}t^4\), when \(t = 2\), we get \( \frac{3}{4}(16) = 12\).- For \(\frac{15}{20}t^5\), when \(t = 2\), we get \(\frac{3}{4}(32) = 12\).- For \(\frac{15}{6}t^3\), when \(t = 2\), we get \(\frac{5}{2}(8) = 20\).Add the results to obtain \(12 + 12 + 20 = 44\).
6Step 6: Conclusion
The value of the integral \(\int_{C} y \, dx + z \, dy + x \, dz\) over the given curve is \(44\).
Key Concepts
Parametrization of CurvesDifferential FormsDefinite Integral
Parametrization of Curves
When dealing with curves and calculating line integrals, it’s essential to understand how to represent these curves using a set of parametric equations. Parametrization allows you to express coordinates of a curve in terms of a single parameter, usually denoted as "t".
In our example, the curve is given in its parametric form as follows:
By expressing each variable in terms of \(t\), we can easily calculate differentials, such as \(dx\), \(dy\), and \(dz\), which we'll need for further calculations. It’s akin to giving us instructions on how to draw the path by changing the value of \(t\).
In our example, the curve is given in its parametric form as follows:
- \( x = 3t \)
- \( y = t^3 \)
- \( z = \frac{5}{4}t^2 \)
By expressing each variable in terms of \(t\), we can easily calculate differentials, such as \(dx\), \(dy\), and \(dz\), which we'll need for further calculations. It’s akin to giving us instructions on how to draw the path by changing the value of \(t\).
Differential Forms
Differential forms play a pivotal role in simplifying the equation of a line integral over a parametrized curve. They essentially break down the complex motion along a curve into understandable pieces by considering infinitesimally small changes.
In the process of evaluating the line integral \( \int_{C} y \, dx + z \, dy + x \, dz \), our primary task is to find the differential forms \(dx\), \(dy\), and \(dz\). Based on our parametric equations, these differentials are:
In the process of evaluating the line integral \( \int_{C} y \, dx + z \, dy + x \, dz \), our primary task is to find the differential forms \(dx\), \(dy\), and \(dz\). Based on our parametric equations, these differentials are:
- \( dx = 3 \, dt \)
- \( dy = 3t^2 \, dt \)
- \( dz = \frac{5}{2}t \, dt \)
Definite Integral
The definite integral ties everything together by accumulating the values obtained from a given interval, making it a powerful tool in calculus to find precise values over curves or surfaces.
After determining the differential forms and setting up the integral \( \int_{0}^{2} (t^3 \, 3) + \left(\frac{5}{4}t^2 \, 3t^2\right) + (3t \, \frac{5}{2}t) \, dt \), the next step involves simplifying and solving this expression term by term. This integral is evaluated over the interval \([0, 2]\), which means it accounts for all the tiny contributions from the entire curve between \(t = 0\) and \(t = 2\).
By integrating each part separately, we find the antiderivatives and use the limits to evaluate the definite integral:
After determining the differential forms and setting up the integral \( \int_{0}^{2} (t^3 \, 3) + \left(\frac{5}{4}t^2 \, 3t^2\right) + (3t \, \frac{5}{2}t) \, dt \), the next step involves simplifying and solving this expression term by term. This integral is evaluated over the interval \([0, 2]\), which means it accounts for all the tiny contributions from the entire curve between \(t = 0\) and \(t = 2\).
By integrating each part separately, we find the antiderivatives and use the limits to evaluate the definite integral:
- The result of the first integral, \(3t^3\), yields \(\frac{3}{4}t^4\).
- For the second, \(\frac{15}{4}t^4\), it simplifies to \(\frac{15}{20}t^5\).
- Lastly, \(\frac{15}{2}t^2\) integrates into \(\frac{15}{6}t^3\).
Other exercises in this chapter
Problem 26
Find the volume of the solid bounded by the graphs of the given equations. $$ z=4-y^{2}, x=3, x=0, y=0, z=0, \text { first octant } $$
View solution Problem 26
In Problems \(25-32\), verify the given identity. Assume continuity of all partial derivatives. $$ \nabla \times(\mathbf{F}+\mathbf{G})=\nabla \times \mathbf{F}
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Find the points on the given surtace at which the tangent plane is parallel to the indicated plane. $$ x^{2}-2 y^{2}-3 z^{2}=33 ; 8 x+4 y+6 z=5 $$
View solution Problem 26
Find the first partial derivatives of the given function. $$ f(x, y)=\frac{x y}{\left(x^{2}-y^{2}\right)^{2}} $$
View solution