# Integral over a curve

## 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?

Simon Bridge
Homework Helper

@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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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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• MapleProb7.png
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Simon Bridge
Homework Helper
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

Ray Vickson
Homework Helper
Dearly Missed
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

Thank you both for your responses. I see now that I was dropping the x^2 and keeping the y'(X) term. *sigh*

Simon Bridge