Closed Curve Integral

**bugatti79** · January 11th 2012, 11:06 AM

Folks,

Find a simple closed curve with counter clockwise rotation that maximizes the value of $\int_{C} \frac{1}{3} y^3 dx + (x-\frac{1}{3} x^3) dx$

If I apply greens theoremI calculate $\int \int_R (1-x^2-y^2) dA= \int_{0}^{2 \pi} \int_{0}^{1} (1-r^2)r dr d \theta = \frac{1}{2}$ ...?

**Ridley** · January 11th 2012, 01:35 PM

Everything looks good except your answer. That integral does not evaluate to 1/2.

You want to maximize $2\pi \int_a^b f'(r)dr=2\pi(f(b)-f(a))$ where $f'(r)=(1-r^2)r$ . Setting $f'(r)=0$ reveals where f has its min and max.

**bugatti79** · January 11th 2012, 10:40 PM

Originally Posted by Ridley

Everything looks good except your answer. That integral does not evaluate to 1/2.

You want to maximize $2\pi \int_a^b f'(r)dr=2\pi(f(b)-f(a))$ where $f'(r)=(1-r^2)r$ . Setting $f'(r)=0$ reveals where f has its min and max.

Sorry, that answer is a typo. It should read pi / 2.

How come we dont calculate a numerical answer even though the limits are specified..

**chisigma** · January 11th 2012, 11:50 PM

Originally Posted by bugatti79

Folks,

Find a simple closed curve with counter clockwise rotation that maximizes the value of $\int_{C} \frac{1}{3} y^3 dx + (x-\frac{1}{3} x^3) dx$

If I apply greens theoremI calculate $\int \int_R (1-x^2-y^2) dA= \int_{0}^{2 \pi} \int_{0}^{1} (1-r^2)r dr d \theta = \frac{1}{2}$ ...?

The request was to find the close curve C that maximizes the integral $\int_{C} f(r)\ dr$ . In Your computation You suppose that C is the unity circle without supplying a prove of that...

Kind regards

$\chi$ $\sigma$

**chisigma** · January 12th 2012, 12:16 AM

Because $x^{2}+y^{2}$ is 'invariant to rotation' we can suppose that C is a circle centered in the origin and with radious R, so that we have to find the R that maximizes the contour integral. We can write...

$\int \int_{A} (1-x^{2}-y^{2})\ dx\ dy = 2\ \pi\ \int_{0}^{R} (1-r^{2})\ r\ dr$ (1)

If we compute the derivative of the (1) respect to R we find that it vanishes for $R\ (1-R^{2})=0$ so that R=1 is the requested value and the curve is [effectively...] the unity circle...

Kind regards

$\chi$ $\sigma$

**bugatti79** · January 12th 2012, 09:27 AM

Originally Posted by chisigma

Because $x^{2}+y^{2}$ is 'invariant to rotation' we can suppose that C is a circle centered in the origin and with radious R, so that we have to find the R that maximizes the contour integral. We can write...

$\int \int_{A} (1-x^{2}-y^{2})\ dx\ dy = 2\ \pi\ \int_{0}^{R} (1-r^{2})\ r\ dr$ (1)

If we compute the derivative of the (1) respect to R we find that it vanishes for $R\ (1-R^{2})=0$ so that R=1 is the requested value and the curve is [effectively...] the unity circle...

Kind regards

$\chi$ $\sigma$

1) Looking at the double integral alone with the limits, without any concern for maximising...the answer is pi/2?

2) Where did you get that term R(1-R^2)

3) I find this conflicting with post #2..

THanks

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