Problem 19
Question
Compute both sides of \(\oint \mathrm{F} \cdot \mathrm{a} d s=\iint\left(M_{x}+N_{y}\right) d x d y\) in 15-20. \(\mathbf{F}=x^{2} y \mathbf{j}\) in the unit triangle (sides \(\left.x=0, y=0, x+y=1\right)\).
Step-by-Step Solution
Verified Answer
Both sides equal: \( \oint \mathbf{F} \cdot ds = \iint (M_x + N_y) dx dy = \frac{1}{12} \).
1Step 1: Define the problem and region
We are tasked with evaluating both sides of the equation \( \oint \mathbf{F} \cdot d\mathbf{s} = \iint (M_x + N_y) dx dy \), where \( \mathbf{F} = x^2 y \mathbf{j} \) over a triangular region with boundaries \( x=0 \), \( y=0 \), and \( x+y=1 \). This is a right triangle with vertices at the origin (0,0), (1,0), and (0,1).
2Step 2: Evaluate the line integral
The line integral around the boundary of the triangle can be split into three parts due to the piecewise linear boundary. Since \( \mathbf{F} = 0 \mathbf{i} + x^2 y \mathbf{j} \), and considering \( \begin{align*} \oint \mathbf{F} \cdot d\mathbf{s} = \int_{C_{1}} \mathbf{F} \cdot d\mathbf{s} + \int_{C_{2}} \mathbf{F} \cdot d\mathbf{s} + \int_{C_{3}} \mathbf{F} \cdot d\mathbf{s} \end{align*} \, we evaluate at each segment: - For segment \( C_1 \) (where \( y=0 \), from (0,0) to (1,0)), \( y=0 \) implies \( x^2 y = 0 \).- For segment \( C_2 \) (where \( x=1 \), from (1,0) to (0,1)), \( x=1 \) so \( \mathbf{F} = y \mathbf{j} \), and since the line is vertical, \( ds = dy \), implying \( 0= C_1 + \int_0^1 y \cdot 1 \cdot dy \), this gives \( \frac{1}{2} \).- In segment \( C_3 \) (where \( x+y=1 \), from (0,1) to (0,0)), the line integral also evaluates to \( -\frac{1}{2} \) after substituting \( y=1-x d = \,x \), dy = -dx \cdot dy \rightarrow \text{as}\, \quad \large{-(-dy)=dx; \,\).
3Step 3: Set up the double integral for the region
The given vector field is \( \mathbf{F} = x^2 y \mathbf{j} \). We note that \( M = 0 \) and \( N = x^2 y \), hence \( N_y = \frac{\partial}{\partial y}(x^2 y) = x^2 \). Thus, - The double integral \( \iint (M_x+N_y) dx dy\) becomes \( \iint x^2 dy dx \).
4Step 4: Evaluate the double integral
The bounds for the double integral correspond to the region of the triangle, which means for each \( x \), \( y \) varies from 0 to \( 1-x \). Hence:\[ \iint x^2 dy dx = \int_{x=0}^{x=1} \left( \int_{y=0}^{y=1-x} x^2 dy \right) dx = \int_{x=0}^{x=1} x^2 (1-x) dx \]. This simplifies to \[ \int_{0}^{1} (x^2 - x^3) dx = \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_{0}^{1} = \left( \frac{1}{3} - \frac{1}{4} \right) \].
5Step 5: Simplify and conclude
Continuing from our previous evaluation, the integral \( \frac{1}{3} - \frac{1}{4} = \frac{4}{12} - \frac{3}{12} = \frac{1}{12} \). The value of the double integral matches the value from the line integral calculated in the previous steps as anticipated by Green's Theorem.
Key Concepts
Line IntegralGreen's TheoremDouble IntegralVector Field
Line Integral
A line integral calculates the total along a curve for a given function, often representing work done by a force field along a path. It's crucial in physics and engineering, where calculating the work along a curve is needed.
When evaluating a line integral for a vector field \( \mathbf{F} \) across a closed curve \( C \), the line integral is expressed as \( \oint \mathbf{F} \cdot d\mathbf{s} \), summing contributions from each infinitesimal segment of the path.
In our problem, the line integral is performed around a triangle with vertices at (0,0), (1,0), and (0,1).
When evaluating a line integral for a vector field \( \mathbf{F} \) across a closed curve \( C \), the line integral is expressed as \( \oint \mathbf{F} \cdot d\mathbf{s} \), summing contributions from each infinitesimal segment of the path.
In our problem, the line integral is performed around a triangle with vertices at (0,0), (1,0), and (0,1).
- For \( C_1 \) from (0,0) to (1,0), the vector field contributes nothing since \( y = 0 \), making \( x^2 y = 0 \).
- For \( C_2 \) from (1,0) to (0,1), where \( x = 1 \), and \( ds = dy \), we compute this part of the integral.
- For \( C_3 \) from (0,1) to (0,0), we account for changes as \( x + y = 1 \), and adapt \( dy \) effectively to get the result.
Green's Theorem
Green's Theorem is a crucial theorem in vector calculus. It connects the microscopic behavior of a field inside a region with the macroscopic behavior along its boundary. Simply put, Green's Theorem relates a line integral around a closed curve \( C \) and a double integral over the plane region enclosed by \( C \).
The theorem is formalized as:
\[ \oint_C \, \mathbf{F} \, \cdot \, d\mathbf{s} = \iint_{D} (M_x + N_y) \, dx \, dy \]where \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \).
In the given problem, the vector field is \( \mathbf{F} = x^2 y \mathbf{j} \), where \( M = 0 \) and \( N = x^2 y \). Calculating \( N_y \), or the partial derivative of \( N \) with respect to \( y \), gives \( x^2 \), leading to our double integral calculation.
Green's Theorem helped us verify our line integration by providing an equivalent area (double) integral over the triangular region, ensuring our solution's consistency.
The theorem is formalized as:
\[ \oint_C \, \mathbf{F} \, \cdot \, d\mathbf{s} = \iint_{D} (M_x + N_y) \, dx \, dy \]where \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \).
In the given problem, the vector field is \( \mathbf{F} = x^2 y \mathbf{j} \), where \( M = 0 \) and \( N = x^2 y \). Calculating \( N_y \), or the partial derivative of \( N \) with respect to \( y \), gives \( x^2 \), leading to our double integral calculation.
Green's Theorem helped us verify our line integration by providing an equivalent area (double) integral over the triangular region, ensuring our solution's consistency.
Double Integral
A double integral extends integration across two dimensions, applicable to areas on a plane. It calculates the 'volume' under a surface, helpful in physical problems involving mass calculation, area, or flux.
For the given triangular region, the double integral translates to the integral of \( x^2 \), represented as:
\[ \iint_D x^2 \, dy \, dx \]
First, the bounds of \( y \) range from 0 to \( 1-x \), while the bounds of \( x \) run from 0 to 1. Thus, the integral simplifies to:
\[ \int_{0}^{1} \left( \int_{0}^{1-x} x^2 \, dy \right) dx = \int_{0}^{1} x^2 (1-x) \, dx \]
Thus, effectively managing these bounds enables reaching the solution: \( \frac{1}{12} \).
Double integral evaluation often requires careful setup of the integration bounds, balancing complexity with precision, especially in non-rectangular regions like our triangle.
For the given triangular region, the double integral translates to the integral of \( x^2 \), represented as:
\[ \iint_D x^2 \, dy \, dx \]
First, the bounds of \( y \) range from 0 to \( 1-x \), while the bounds of \( x \) run from 0 to 1. Thus, the integral simplifies to:
\[ \int_{0}^{1} \left( \int_{0}^{1-x} x^2 \, dy \right) dx = \int_{0}^{1} x^2 (1-x) \, dx \]
Thus, effectively managing these bounds enables reaching the solution: \( \frac{1}{12} \).
Double integral evaluation often requires careful setup of the integration bounds, balancing complexity with precision, especially in non-rectangular regions like our triangle.
Vector Field
A vector field assigns a vector to each point in space. Commonly visualized as a collection of arrows, where direction and length vary geographically, vector fields display real-world phenomena.
Examples include gravitational fields, fluid dynamics, and in our case, the field \( \mathbf{F} = x^2 y \mathbf{j} \), which has components \( x^2 y \mathbf{j} \), meaning its "force" acts purely in the \( y \)-direction with magnitude dependent on \( x^2 y \).
The interpretation of a vector field helps students to visualize how particles, like fluids, would flow through the region. In our problem, the effect is bounded by the triangular area, showing flow characteristics determined by our integral paths (edges of the triangle).
Examples include gravitational fields, fluid dynamics, and in our case, the field \( \mathbf{F} = x^2 y \mathbf{j} \), which has components \( x^2 y \mathbf{j} \), meaning its "force" acts purely in the \( y \)-direction with magnitude dependent on \( x^2 y \).
The interpretation of a vector field helps students to visualize how particles, like fluids, would flow through the region. In our problem, the effect is bounded by the triangular area, showing flow characteristics determined by our integral paths (edges of the triangle).
- The field dictates how each segment of our path contributes to the integral.
- It characterizes how forces interact along or within the defined area.
Other exercises in this chapter
Problem 19
Find a potential \(f\) if it exists. $$\mathbf{F}=z \mathbf{i}+\mathbf{j}+x \mathbf{k}$$
View solution Problem 19
Describe the closed surface \(S\) and outward normal \(n\) : (a) \(V=\) hollow ball \(1 \leqslant x^{2}+y^{2}+z^{2} \leqslant 9\) (b) \(V=\) solid cylinder \(x^
View solution Problem 19
\(\mathrm{~A}\) wire hoop around a vertical circle \(x^{2}+z^{2}=a^{2}\) has density \(\rho=a+z\). Find its mass \(M=\int \rho d s\).
View solution Problem 20
Find a potential \(f\) if it exists. $$\mathbf{F}=2 x y z \mathbf{i}+x^{2} z \mathbf{j}+x^{2} y \mathbf{k}$$
View solution