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June 25, 2016
Posted by **ms** on Wednesday, September 18, 2013 at 4:19am.

I have gone thus- (theta written as t)

r^2= a^2 sin^6 t/3 and

(dr/dt)^2=a^2 sin^4(t/3)cos^2(t/3)

s=Int Sqrt[a^2 sin^6 t/3+a^2 sin^4(t/3)cos^2(t/3)]dt

=a Int Sqrt[sin^4(t/3){(sin^2(t/3)+cos^2(t/3)}]dt=a Int Sqrt[sin^4(t/3)dt

=a Int sin^2(t/3)dt

=a[1/2(t/3)-1/2{3sin(2t/3)}] from t=0 to 2pi

Or, s=a.pi/3-3a[sin(4pi/3)]/4

=a.pi/3+3a[sin(pi/3)]/4

=a.pi/3+3/4*rt3/2=a.pi/3+3rt3/8.

The answer in the book is 3a.pi/2. Where have I gone wrong?

- calculus -
**Graham**, Wednesday, September 18, 2013 at 6:39amr = a sin^3(θ/3)

r' = a sin^2(θ/3) cos(θ/3)

Sketch the curve. sin^2(θ/3) has a period of 3π.

The complete length of r must be measured over one whole period; thus integrate from 0 to 3π, (not 2π).

Arc length in polar coordinates:

s = ∫{0↔3π} √(r^2 + r'^2) dθ

s = ∫{0↔3π} √(a^2 sin^6(θ/3) + a^2 sin^4(θ/3)cos^2(θ/3)) dθ

s = a ∫{0↔3π} sin^2(θ/3) √(sin^2(θ/3) + cos^2(θ/3)) dθ

s = a ∫{0↔3π} sin^2(θ/3) dθ

Use sin^2(A) = (1-cos(2A))/2

s = (a/2) ∫{0↔3π} 1 - cos(2θ/3) dθ

s = (a/2) [θ - (3/2) sin(2θ/3)]{θ=0↔3π}

s = (a/2) (3π - (3/2) sin(4π))

s = 3aπ/2

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Error 1: The period over which you integrate

Error 2: The integration of sin^2(θ/3). - calculus -
**ms**, Wednesday, September 18, 2013 at 7:01amThanks a lot for guiding. Kindly let me know if it should be a routine procedure to plot the complete curve everytime or can it be conveniently worked out analytically?

I used this formula for integrating Int sin^2x dx=1/2*(x-1/2*sin2x)which is a std formula. Why the result differs? - calculus -
**Graham**, Wednesday, September 18, 2013 at 7:53amYes, the general formula is.

∫sin^2(x) dx = x/2 - sin(2x)/4 +C

However, a more general formula is:

∫sin^2(ax) dx = x/2 - sin(2ax)/4a +C

Proof:

Let I = ∫ sin^2(ax) dx

Substitute u = ax, and dx = (1/a)du

I = (1/a)∫sin^2(u) du

Use your standard formula

I = (1/a)(u/2 - sin(2u)/4) +C

Substitute back, u = ax

I = (1/a)(ax/2 - sin(2ax)/4) +C

Simplify and conclude:

.: ∫ sin^2(ax) dx = x/2 - sin(2ax)/4 +C

Q.E.D.

Basically; when you substitute for x, you must remember to also substitute for dx. - calculus -
**ms**, Wednesday, September 18, 2013 at 8:02amMr. Graham,

Thank you very much for elaborate explanation. Kindly also let me know if it is ok to find at what values of theta the r becomes 0 and take that as period of curve? If so, plotting the complete curve every time may not be required. - calculus -
**Graham**, Wednesday, September 18, 2013 at 8:21amFor a closed curve, the complete length is the distance along the curve from a starting point, and to the starting point again. That is, measure the length of a complete circuit.

When using polar coordinates, the domain over which you need to integrate to measure the complete length is called the period. The period being the rotation you need to traverse to trace the curve.

This is often, but not necessarily 2π. - calculus -
**ms**, Thursday, September 19, 2013 at 12:04amThank you very much.