polar coordinates...help again please... Watch

Schmeevey
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#1
Report Thread starter 12 years ago
#1
i literally am spending hours on these two questions.

1) find the finite region under the half line with eqn

f(theta) = a/(1 + costheta) theta1=0 theta2= pi/2

where r=f(theta) etc...

so far I have done...

integral 1/2 r^2 dtheta = 1/2a^2 integral (1+costheta)^2 dtheta

= 1/2a^2(-(1+costheta)^-1)

putting theta values in, i get:

1/2a^2(-1/(1+cos90 - -1/1+cos0)
=1/2a^2(-1--0.5)
=-1/4a^2

BUT the answer in the back of the book says it is 1/3a^2.......how??

2) find the finite region under the half line with eqn r^2cos2theta=a^2
theta1 = pi/6 theta2 = -pi/6

so far i have done...

1/2r^2 dtheta = integral 1/2 (a^2/cos2theta) dtheta

= 1/2a^2 integral sec2theta dtheta

= 1/2a^2(In|sec2theta + tan2theta|)

putting theta values in gives...

1/2a^2 (In|sec pi/3 + tan pi/3| - In|sec -pi/3 + tan -pi/3|)

=1/2a^2 (In|2+root3| - In|2-root3|)

but the answer is just 1/2a^2(In|2+root3|)

is there some method of cancelling down which i cnt see???

thanks a lotxxx
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steve10
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#2
Report 12 years ago
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For (1), I get (1/3)a² as well.
You didn't integrate the 1/(1+cost)² correctly. Try differentiating your result to see why.
You will have to use the substituion u = tan(t/2). Then it works out.
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steve10
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Report 12 years ago
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For Q. (2), the integration of sec(2t) is again wrong. This time use the substitution u = tan(t). And persevere! It takes a bit of working out.
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Kolya
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Report 12 years ago
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See this thread.
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