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Chapter 15: Problem 14

Let \(C\) be the boundary of the region enclosed between \(y=x^{2}\) and \(y=2 x\). Assuming that \(C\) is oriented counterclockwise, evaluate the following integrals by Green's Theorem: $$ \text { (a) } \oint_{C}\left(6 x y-y^{2}\right) d x \quad \text { (b) } \oint_{C}\left(6 x y-y^{2}\right) d y $$

Short Answer

Expert verified
(a) \(\frac{8}{3}\), (b) \(\frac{64}{5}\)

Step by step solution

01

Identify the Curves and Points of Intersection

First, we identify the curves \(y = x^2\) and \(y = 2x\). To find the points of intersection between these curves, set them equal to each other: \(x^2 = 2x\). Solving for \(x\), we get \(x(x - 2) = 0\), which gives solutions \(x = 0\) and \(x = 2\). Thus, the curves intersect at the points \((0, 0)\) and \((2, 4)\).
02

Define the Region R

The region \(R\) is enclosed by the curves \(y = x^2\) and \(y = 2x\) between the points of intersection \((0,0)\) and \((2,4)\). For \(x\) in \([0, 2]\), \(y\) on \(y=x^2\) will be lower, and \(y\) on \(y=2x\) will be upper.
03

Green's Theorem Setup for Integral (a)

Use Green's Theorem for \(\oint_C P \, dx + Q \, dy = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA\). For part (a), let \(P = 6xy - y^2\) and \(Q = 0\). So, we need \(\frac{\partial Q}{\partial x} = 0\) and \(\frac{\partial P}{\partial y} = 6x - 2y\).
04

Evaluate the Double Integral for (a)

Now, calculate the double integral \(\iint_R (0 - (6x - 2y)) \, dA = \iint_R (2y - 6x) \, dA\). The region is bounded by \(y = x^2\) and \(y = 2x\), so the limits for the integral are \(x = 0\) to \(x = 2\) and \(y = x^2\) to \(y = 2x\). Therefore, integrate \((2y - 6x)\) with respect to \(y\) first, and then \(x\): \[\int_0^2 \int_{x^2}^{2x} (2y - 6x) \, dy \, dx\]
05

Integrate with Respect to y for (a)

First, perform the integral with respect to \(y\): \[\int_{x^2}^{2x} (2y) \, dy = \left[ y^2 \right]_{x^2}^{2x} = (2x)^2 - (x^2)^2 = 4x^2 - x^4\]\[\int_{x^2}^{2x} (-6x) \, dy = -6x(2x - x^2) = -6x(2x - x^2)\]Combine these results:
06

Simplify and Integrate with Respect to x for (a)

Bringing it all together, the expression simplifies to: \[\int_0^2 (4x^2 - x^4 - 6x(2x - x^2)) \, dx = \int_0^2 (4x^2 - x^4 - 12x^2 + 6x^3) \, dx\]Simplify to:\[\int_0^2 (-8x^2 + 6x^3) \, dx\]Now, solve this integral:\[\int_0^2 (6x^3 - 8x^2) \, dx = \left[ \frac{6}{4}x^4 - \frac{8}{3}x^3 \right]_0^2 = \left[ \frac{3}{2}x^4 - \frac{8}{3}x^3 \right]_0^2\]Evaluating gives:\[= \frac{3}{2}(16) - \frac{8}{3}(8) = 24 - \frac{64}{3} = \frac{72}{3} - \frac{64}{3} = \frac{8}{3}\]
07

Green's Theorem Setup for Integral (b)

For part (b), set \(P = 0\) and \(Q = 6xy - y^2\). Thus, we have \(\frac{\partial Q}{\partial x} = 6y\) and \(\frac{\partial P}{\partial y} = 0\). The double integral becomes \(\iint_R (6y - 0) \, dA\).
08

Evaluate Double Integral for (b)

We integrate \(\iint_R 6y \, dA = 6 \iint_R y \, dA\). Using the same bounds \(x = 0\) to \(x = 2\) and \(y = x^2\) to \(y = 2x\):\[6 \int_0^2 \int_{x^2}^{2x} y \, dy \, dx = 6 \int_0^2 \left[ \frac{y^2}{2} \right]_{x^2}^{2x} \, dx\]
09

Integrate with Respect to y and Simplify for (b)

First, integrate with respect to \(y\): \[6 \int_0^2 \left( \frac{(2x)^2}{2} - \frac{(x^2)^2}{2} \right) \, dx = 6 \int_0^2 (2x^2 - \frac{x^4}{2}) \, dx\]Simplify:\[6 \int_0^2 (2x^2 - \frac{x^4}{2}) \, dx = 6 \int_0^2 (2x^2 - \frac{x^4}{2}) \, dx\]
10

Integrate with Respect to x for (b)

Now, integrate with respect to \(x\):\[6 \left[ \frac{2x^3}{3} - \frac{x^5}{10} \right]_0^2 = 6 \left( \frac{16}{3} - \frac{32}{10} \right)\]Simplify:\[= 6 \left( \frac{16}{3} - \frac{16}{5} \right) = 6 \left( \frac{80 - 48}{15} \right) = 6 \left( \frac{32}{15} \right) = \frac{192}{15} = \frac{64}{5}\]
11

Conclusion

By applying Green's Theorem, the values obtained for the line integrals around \(C\) are:(a) \(\oint_C (6xy - y^2) \, dx = \frac{8}{3}\)(b) \(\oint_C (6xy - y^2) \, dy = \frac{64}{5}\)

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Key Concepts

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

Line Integrals
A line integral, also known as a path integral, is a fundamental concept in vector calculus. It is used to integrate functions along a curve or path in the plane or space. In other words, we are measuring something along a curve.
  • In our exercise, we are asked to evaluate line integrals around a closed curve, which is the boundary of a region between two curves.
  • Line integrals can be used to find work done by a force field in moving along a path, among other applications.
  • The integrals \(\oint_C (6xy - y^2) \, dx\) and \(\oint_C (6xy - y^2) \, dy\) represent line integrals of different vector fields around the closed curve \(C\).
To evaluate these integrals, we use Green's Theorem, a key tool that converts line integrals into double integrals, simplifying calculations. By applying it, we move from integrating over a path to integrating over the region the path encloses. This not only simplifies the computation but also provides deeper insights into the nature of the function over the enclosed area.
Double Integrals
Double integrals extend the concept of a regular integral to two dimensions, allowing us to compute volumes under surfaces or areas over regions. In our exercise, double integrals help us calculate the area enclosed by the curves and the values of expressions over this region.
  • We switch our line integral values into double integrals using Green's Theorem, allowing us to evaluate over the region rather than the boundary.
  • The region \(R\) is enclosed by \(y = x^2\) and \(y = 2x\) from x = 0 to x = 2, reflecting where the curves intersect.
  • We set up our double integral limits based on these bounds, integrating first with respect to y, and then x, or vice versa based on convenience of simplification.
Once the change of integration type is made, care must be taken to correctly apply limits and solve step by step. Calculating \(\iint_R (2y - 6x)\, dA\) and \(\iint_R 6y \, dA\) leads us to the required results specified in the exercise: \frac{8}{3}\ for part (a) and \frac{64}{5}\ for part (b).
Calculus Problem Solving
In solving calculus problems involving Green's Theorem as in this exercise, breaking the problem into manageable steps is crucial:
  • Identify the curves and their intersection points; here, \x = 0\ and \x = 2\ were determined through solving \(x^2 = 2x\).
  • Understand the geometric interpretation to properly define the region of integration \(R\).
  • Apply Green's Theorem to convert line integrals into double integrals over \(R\), specifying the functions of interest.
  • Set up correct integral bounds and order of integration, converting into a form that's possible to solve analytically.
  • Perform the integrals carefully, checking adjustments for simplification after each integration step.
By following these systematic steps, tackling complex calculus problems becomes more approachable and understandable, ensuring accuracy and deeper learning. Green's Theorem not only aids in simplification of integration problems but also enhances understanding of vector fields and area computations in calculus.

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