Integral of (x+y)dx+x^2dy over semicircle (y>=0) using Green's Theorem

In summary, the parametric equation of a circle can be expressed in terms of x = 2cos^2(theta), y = 2cos(theta)sin(theta). To evaluate the integral of {(x+y)dx+x^2dy} along the semicircle for which y >=0 from (0,0) to (2,0), the attempt was made to use Green's Theorem, polar coordinates, trig substitution, integration by parts, change of variables, or a combination of these. However, none of these techniques proved successful in obtaining a valid answer. After trying every possible approach, the only thing that ended up yielding a result was using the vector calculus package in Maple to directly evaluate
  • #1
BigFlorida
41
1

Homework Statement


The parametric equations of a circle, center (1,0) and radius 1, can be expressed as x = 2cos^2(theta), y = 2cos(theta)sin(theta).
Evaluate the integral of {(x+y)dx+x^2dy} along the semicircle for which y >=0 from (0,0) to (2,0).

Homework Equations


Perhaps Green's Theorem?

The Attempt at a Solution


I tried using Green's Theorem, I tried using polar coordinates, I tried using a combination of Green's Theorem and polar coordinates, I tried trig substitution, I tried u substitution, I tried integration by parts, I tried change of variables. I literally tried every technique I know for integration except partial fractions. I am at a complete loss. Any thoughts on the setup?
 
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  • #2
Please show working with reasoning for your best attempt.
 
  • #3
@Simon Bridge Here is my cleanest attempt. I got -Pi/2 as my answer, the book says the answer is 2 - Pi/2.
 

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  • #4
Also, I evaluated it using the vector definition of a line integral using Maple, and got Pi as the answer using the parameters and -Pi/2 as the answer without...
 

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  • #5
Green's theorem looks like its complicating things... did you try just going through the usual parameterization stuff first?
You have an integral of the form ##\int_C Pdx + Qdy : P=x+y,\; Q=x^2##
The parameterization of the curve is ##x(t)=2\cos^2t,\; y(t)=2\cos t \sin t##

So the integral becomes:

Check the limits ... you want x to go from 0 to 2.
http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtII.aspx
 
  • #6
BigFlorida said:
Also, I evaluated it using the vector definition of a line integral using Maple, and got Pi as the answer using the parameters and -Pi/2 as the answer without...

When I use Maple 11 to evaluate the integral ##J = \int_0^2 (x + y + x^2 y') \, dx## (with ##y = \sqrt{2x - x^2}##) I get ##J = 2 - \pi/2##, which is the book's answer. I did not bother to use the "VectorCalculus" package, because just doing a direct evaluation seemed much faster and easier.

The images you posted are too small and too blurry to make out, so I am not able to see any of the details of your computations. I would suggest copying and pasting individual lines, rather than attaching an image. For example, here is the indefinite integral (after putting y = sqrt(2*x - x^2) and dy = diff(y,x)):

> int(x+y+x^2*dy, x); lprint(%);
(1/2)*x^2-(1/4)*(2-2*x)*(2*x-x^2)^(1/2)-(1/2)*arcsin(-1+x)+(1/3)*x^2*(2*x-x^2)^(1/2)+(1/3)*x*(2*x-x^2)^(1/2)+(2*x-x^2)^(1/2)

This is copied and pasted directly from the Maple 11 worksheet
 
  • #7
Thank you both for your responses. I see now that I was dropping the x^2 and keeping the y'(X) term. *sigh*
 
  • #8
Well done - always double check the transcription :)
You'll find a lot of responders here are impatient with images - best to cut and paste from text editors and terminals, and learn to use LaTeX markup for the handwritten stuff. The approach is good discipline, forces you to pay attention to what you are doing, and is good practise for later.

Enjoy.
 

Related to Integral of (x+y)dx+x^2dy over semicircle (y>=0) using Green's Theorem

1. What is the definition of an integral over a curve?

An integral over a curve is a mathematical concept that represents the total accumulation of a function along a given curve. It is denoted by ∫Cf(x,y) ds, where C is the curve, f(x,y) is the function being integrated, and ds represents an infinitesimal element of the curve's length.

2. How is an integral over a curve different from a regular integral?

An integral over a curve is different from a regular integral because it involves integrating a function along a specific path or curve, rather than over a fixed interval. This means that the limits of integration for an integral over a curve are not constant, but depend on the shape and location of the curve.

3. What is the significance of an integral over a curve in physics?

In physics, an integral over a curve is used to calculate important physical quantities such as work, force, and energy. It allows us to find the total amount of these quantities along a given path, rather than just at specific points. This is particularly useful in calculating the work done by a force that varies along a curved path.

4. How is an integral over a curve evaluated?

An integral over a curve can be evaluated using various techniques such as the Fundamental Theorem of Calculus, Green's Theorem, or Stoke's Theorem. The method used will depend on the complexity of the curve and the function being integrated. In some cases, numerical methods may also be used to approximate the value of the integral.

5. Can an integral over a curve be negative?

Yes, an integral over a curve can be negative. This occurs when the function being integrated has negative values along certain parts of the curve, resulting in a net accumulation of negative values. The sign of the integral over a curve depends on the direction of traversal along the curve, with a positive direction resulting in a positive integral and a negative direction resulting in a negative integral.

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