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Chapter 10: Problem 3

Using Green's theorem, evaluate \(\int_{c} \mathbf{F}(\mathbf{r}) \cdot d \mathbf{r}\) counterclockwise around the boundary curve \(C\) of the region \(R\), where $$\mathbf{F}=\left[-y^{3}, \quad x^{3}\right], C \text { the circle } x^{2}+y^{2}=25$$

Short Answer

Expert verified
The value of the integral is \( 937.5\pi \).

Step by step solution

01

Understanding Green's Theorem

Green's Theorem relates a line integral around a simple, closed curve to a double integral over the region it encloses. It states that \( \oint_C \mathbf{F} \cdot d \mathbf{r} = \int \int_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \), where \( \mathbf{F} = [M, N] \). For \( \mathbf{F} = [-y^3, x^3] \), \( M = -y^3 \) and \( N = x^3 \).
02

Calculate Partial Derivatives

Compute \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \): \( \frac{\partial N}{\partial x} = \frac{\partial (x^3)}{\partial x} = 3x^2 \) and \( \frac{\partial M}{\partial y} = \frac{\partial (-y^3)}{\partial y} = -3y^2 \).
03

Set Up Double Integral

Substitute the partial derivatives into Green's Theorem: \( \int \int_R \left( 3x^2 - (-3y^2) \right) \, dA = \int \int_R (3x^2 + 3y^2) \, dA \).
04

Transform to Polar Coordinates

Since \( R \) is a circle with radius 5, use polar coordinates: \( x = r \cos \theta \) and \( y = r \sin \theta \). Then \( x^2 + y^2 = r^2 \), hence \( \int \int_R 3(x^2 + y^2) \, dA = \int_0^{2\pi} \int_0^5 3r^2 \cdot r \, dr \, d\theta = 3 \int_0^{2\pi} \int_0^5 r^3 \, dr \, d\theta \).
05

Evaluate Inner Integral

Evaluate \( \int_0^5 r^3 \, dr \): \( \int r^3 \, dr = \frac{r^4}{4} \), thus \( \left[ \frac{r^4}{4} \right]_0^5 = \frac{625}{4} \).
06

Evaluate Outer Integral

Substitute the result from the previous step into the integral: \( 3 \int_0^{2\pi} \frac{625}{4} \, d\theta = \frac{1875}{4} \int_0^{2\pi} \, d\theta \). Compute \( \int_0^{2\pi} \, d\theta = 2\pi \).
07

Compute Final Answer

Evaluate the final product: \( \frac{1875}{4} \times 2\pi = \frac{3750}{4} \pi = 937.5\pi \).

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Key Concepts

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

Line Integrals
Line integrals are a type of integral where you sum up values along a curve. It鈥檚 like adding up tiny pieces over a path. When dealing with vector fields, line integrals help us find physical quantities like work done by a force field around a path.
  • Consider a vector field \( \mathbf{F} \), which includes forces represented as vectors. A line integral involves calculating the contribution of \( \mathbf{F} \) along a curve \( C \).
  • Mathematically, we express a line integral as \( \int_C \mathbf{F} \cdot d\mathbf{r} \). Here, \( d\mathbf{r} \) represents a tiny step along the curve \( C \).
  • In practical terms, think of it as measuring the accumulated push of a vector field along a path.
Green's Theorem connects these integrals to those over an area, allowing for simpler computation.
Double Integrals
Double integrals extend the concept of integration to two dimensions, typically over a region in the plane. They help calculate volume under a surface or more abstract quantities like total mass or charge.
  • When you compute a double integral \( \int \int_R f(x, y) \, dA \), you're summing up small patches, \( dA \), over the region \( R \). The function \( f(x, y) \) tells you the height or value over each patch.
  • Double integrals can be converted into more manageable forms using polar coordinates, especially useful for circular regions. In polar, the area element \( dA \) becomes \( r \, dr \, d\theta \).
  • In the given exercise, converting to polar coordinates simplified evaluating the integral over a circle.
By using double integrals, complex two-dimensional analysis, like evaluating areas enclosed by curves, becomes more feasible.
Partial Derivatives
Partial derivatives measure how a function changes as one of its variables change, keeping others constant. They're essential in multivariable calculus to understand how functions behave.
  • Given a function of two variables, such as \( f(x, y) \), partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) indicate the rate of change of the function with respect to \( x \) and \( y \) separately.
  • In Green鈥檚 Theorem, partial derivatives help in transforming a line integral into a double integral by identifying changes in vector components: \( M \) with respect to \( y \) and \( N \) with respect to \( x \).
  • In our solution, \( M = -y^3 \) and \( N = x^3 \), so \( \frac{\partial N}{\partial x} = 3x^2 \) and \( \frac{\partial M}{\partial y} = -3y^2 \). These values were crucial for setting up the double integral.
Partial derivatives are vital for the transition from single-variable calculus to multidimensional calculus, helping to solve real-world problems involving surfaces and volumes.

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Most popular questions from this chapter

PROJECT. Tangent Planes T(P) will be less Importunt in our work, but you should linosw bow to represent them. (a) If \(S: r(u, v),\) then \(T(P):\left(x^{*}-r \quad \mathbf{r}_{u} \quad \mathbf{r}_{v}\right)=0\) (a scalar triple product) of \(\mathbf{r}^{*}(p, q)-\mathbf{r}(P)+p \mathbf{r}_{\mathbf{w}}(P)+q \mathbf{r}_{r}(P)\) (b) If \(S: g(x, y, z)=0,\) then \(T(P):\left(r^{*}-r(P)\right) \cdot \nabla_{8}=0\) (c) If \(S: z=f(x, y)\), then \(\left.T(P): z^{*}-z=\left(x^{*}-x\right) f_{x}(P)+\left(y^{*}-y\right) f_{y}(P)\right)\) Interpret (a)-(c) geometrically. Give rwo examples for (a), two for (b), and two for (c).

Find the volume of the following regions in space The tetrahedron cut from the first octant by the plane \(^{3}_{2} x+2 y+z=6 .\) Check by vector methods.

Evaluate the integral \(\iint_{S}(\operatorname{curl} \mathbf{F}) \cdot \mathbf{n} d A\) directly for the given \(\mathbf{F}\) and \(\mathbf{S}\). $$\begin{aligned}&\mathbf{F}=\left[\begin{array}{lll}x^{2} & 2 & 2 \end{array}\right]\\\&S \text { the square } 0 \leq x \leqq a, 0 \leqq y \leqq a, z=1\end{aligned}$$

Using Green's theorem, evaluate \(\int_{c} \mathbf{F}(\mathbf{r}) \cdot d \mathbf{r}\) counterclockwise around the boundary curve \(C\) of the region \(R\), where $$\begin{aligned} &\mathbf{F}=\left[\begin{array}{ll} -e^{3}, & e^{\pi} \end{array}\right], R \text { the triangle with vertices }(0,0),\\\ &(2,0),(2,1) \end{aligned}$$

Evaluate this integral by the divergence theorem. (Show the details.) \(\mathbf{F}=\left[\begin{array}{lll}5 x^{2}, & 5 y^{2}, & 5 z^{3}\end{array}\right], S \cdot x^{2}+y^{2}+z^{2}=4\)
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