Problem 19 - The Calculus with Analytic Geometry

Question

\(\int_{C}(2 x \cos y-3) d x-\left(x^{2} \sin y+z^{2}\right) d y-(2 y z-2) d z ; A\) is \((-1,0,3)\) and \(B\) is \((1, \pi, 0) ;\) Exercise 9

Step-by-Step Solution

Answer

\( \int_{C}(2 x \cos y-3) dx- (x^2 \sin y+z^2) dy-(2 y z-2) dz= -6-\pi\int_0^1(1-16t+20t^2 dt)=-6-4= -10 \).

Step 1: Understand the Problem

We are given a line integral of a vector field over a curve C. The vector field is \(F = (2x\cos y - 3, - (x^2 \sin y + z^2), - (2y z - 2)) \). Points A and B are given as (-1,0,3) and (1,\pi,0) respectively.

Step 2: Parameterize the Curve

We parameterize the straight line from A to B using a parameter t where 0 \leq t \leq 1. The parameterization is: \( x(t) = -1 + 2t \, y(t) = \pi t \, z(t) = 3 - 3t \).

Step 3: Compute Derivative of Parameterization

Find the derivatives of the parameterized components with respect to t: \( dx/dt = 2;\, dy/dt = \pi; \, dz/dt = -3 \).

Step 4: Substitute Parameterization into Vector Field

Substitute \( x(t), y(t) \) and \( z(t) \) into the vector field components to get \( F(x(t), y(t), z(t)) \). We get: \( F((2(-1+2t)\cos(\pi t)-3, \ -( (-1+2t)^2 \sin(\pi t)+(3-3t)^2 ), \ -(2(\pi t)(3 - 3t) - 2)) \).

Step 5: Dot Product with Derivative

Take the dot product of F with the derivatives of the parameterization: \( (2(-1+2t)\cos(\pi t)-3) 2 - ( (-1+2t)^2 \sin(\pi t)+(3-3t)^2 ) \pi - (2(\pi t)(3 - 3t) - 2)(-3) \), which simplifies to: \ 4(-1+2t)\cos(\pi t) - 6 \ - \pi((-1+2t)^2 \sin(\pi t) + (3- 3t)^2) +3(2\pi t(3-3t) -2) \.

Step 6: Integrate with Respect to t

Integrate the above expression with respect to t from 0 to 1: \( \int_{0}^{1} (4(-1+2t)\cos(\pi t) - 6 \ - \pi((-1+2t)^2 \sin(\pi t) + (3- 3t)^2) +3(2\pi t(3-3t) -2)) dt \). Perform the integration carefully considering each term.

Step 7: Evaluate the Integral

Compute the definite integrals of each part separately and sum the results.

Key Concepts

Vector Field

A vector field assigns a vector to every point in space. In this exercise, we have a vector field \(\textbf{F} = (2x \cos y - 3, - (x^{2} \sin y + z^{2}), - (2y z - 2))\). Each term represents a component of the field along one of the coordinate axes.

Visualize the vector field as arrows spread throughout space, each showing the direction and magnitude of the vector at that point. Understanding this graphic helps when dealing with line integrals, as it informs how the field interacts with the curve.

• The first component, \(2x \cos y - 3\), depends on variables x and y.
• The second component, - (x^{2} \sin y + z^{2}), relies on all three variables x, y, and z.
• The third component, - (2y z - 2), depends on variables y and z.
  
  Visualizing the components interfering with the path of the curve will aid in grasping their integrals.

Parameterization

Parameterization converts a curve or line into a one-parameter function, facilitating calculations along that path.

To parameterize the straight line from A (-1,0,3) to B (1, \pi, 0), we introduce a parameter \ t that varies from 0 to 1:
\( x(t) = -1 + 2t \)
\( y(t) = \pi t \)
\( z(t) = 3 - 3t \)

• At \ t = 0, the coordinates are \(-1, 0, 3\), corresponding to point A.
• At \ t = 1, the coordinates are \(1, \pi, 0\), corresponding to point B.
  With this parameterization, we transform a 3D spatial integration problem into a 1D problem over t.

Definite Integral

A definite integral calculates the accumulation of quantities, and here, it sums up values of a vector field along a specified curve.

The line integral given in the problem is:
\( \int_{C}(2 x \cos y-3) d x-\left(x^{2} \sin y+z^{2}\right) d y-(2 y z-2) d z \)
To compute this, we use parameterization to express integrals in terms of the parameter \ t:
\( \int_{0}^{1} (4(-1+2t)\cos(\pi t) - 6 - \pi((-1+2t)^{2}\sin(\pi t) + (3-3t)^{2}) + 3(2\pi t(3-3t) - 2)) dt \)
This expression needs integration over \ t from 0 to 1, simplifying calculations by focusing on changes over one variable.

Dot Product

The dot product of two vectors produces a scalar and measures the extent to which they point in the same direction.

In the context of line integrals, the dot product is used to combine the vector field with the derivatives of parameterization:
We have the vector field components \( F(x(t), y(t), z(t)) = (2(-1+2t)\cos(\pi t)-3, - ((-1+2t)^{2} \sin(\pi t) + (3-3t)^{2}), - (2(\pi t)(3 - 3t) - 2)) \).
The derivatives are

• dx/dt = 2
• dy/dt = \pi
• dz/dt = -3
  
  
  The dot product of these with the vector field gives the integrand for our definite integral:
  \( (2(-1+2t)\cos(\pi t)-3) \cdot 2 - (( -1+2t)^{2}\sin(\pi t)+(3-3t)^{2} ) \cdot \pi - (2(\pi t)(3-3t) - 2) \cdot (-3) \).
  The resulting scalar function then integrates over the parameter t.