Using an integral to find area of section of a sphere

[Azelketh]

Homework Statement

(sorry i have not learned latex to display the maths)
Integrating to find the proportion of the area of a sphere that is within the section described by the bounds of theta.
theta is longitude axis angle
-(pi)/9 <= theta <= pi/9
thi is lattitudal axis angle
0 <= thi <= 2 pi

as using the +pi/9 and -pi/9 limits for integration would return 0,
i have used 0 to pi/9 and then multiplied the result by 2.

integral( 0 to pi/9) integral(0 to 2 pi) { 2 r^2 sin(theta) d thi d theta }

= integral( 0 to pi/9) { 4 pi r^2 sin(theta) d theta }

= 4 pi r^2 [-cos(pi/9) - cos(0)]

= 4 pi r^2 [-cos(pi/9) - 1] = 0.06... *(4 pi r^2)

this must be incorrect however as the fraction of the surface of the sphere should surely be fraction of the angle. eg the top half of the sphere should have 1/2 the surface area, so if the range of theta is 2pi/9 surely the area of the sphere should be (4 pi r^2)* [ ( 2 pi / 9 ) /pi]
should therefore be 8 pi r^2 / 9

can anyone point out where i am going wrong in the integral or my thinking?

 

[issacnewton]

your area element doesn't look correct

 

[Azelketh]

area element = r^2 sin(theta) d theta d thi
is this more correct? with the order of the d thi and d theta reversed? this shouldent make any difference though if the correct set of limits for the integrals are used.

 

[jackmell]

You may wish to consider computing the area via surface of revolution by revolving the segment of the curve $x^2+y^2=r^2$ around the x-axis between $(\pi/2-\pi/9, \pi/2+\pi/9)$

Just seems easier that way.

 

[Azelketh]

problem solved, it was just the integral limits which were the wrong way around, also as cos is an even function istead of integrating over 0 to 2 pi which would return 0, integrating from 0 to pi and doubling the result gives the correct answer.
Thanks for your suggestions dudes.