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1. Sketch, in the same diagram, the curves with equations r=3cosθ and r=1+cosθ and find the area of the region lying inside both curves.
I get the integral(s) required for the area to be that attached, and evaluate it to get 7pi/2. However, the book gives the answer as 5pi/4.
Am i making a mistake somewhere? Thanks.
Attached Images

2. (Original post by Gaz031)
Sketch, in the same diagram, the curves with equations r=3cosθ and r=1+cosθ and find the area of the region lying inside both curves.
I get the integral(s) required for the area to be that attached, and evaluate it to get 7pi/2. However, the book gives the answer as 5pi/4.
Am i making a mistake somewhere? Thanks.
From the integral that you've derived, 7pi/2 is defintely the right answer.
Either the book is wrong, which often does happen, or you need to consider and check your integral(s) again, or even the limits of the integrals.
3. no - i'm getting 15 pi/12 = 5pi/4 a^2

check your integration - i think your limits are on the wrong equations

i got limits of pi to pi/3 for the (1+cos)^2 eqtn and between pi/3 and 0 for the 3cos eqtn

phil
4. (Original post by Gaz031)
Sketch, in the same diagram, the curves with equations r=3cosθ and r=1+cosθ and find the area of the region lying inside both curves.
I get the integral(s) required for the area to be that attached, and evaluate it to get 7pi/2. However, the book gives the answer as 5pi/4.
Am i making a mistake somewhere? Thanks.
my book has answer of 5pi/8 a^2 (this is qeutsion 37 b in rewiew exercise isn't it?
5. (Original post by Phil23)
my book has answer of 5pi/8 a^2 (this is qeutsion 37 b in rewiew exercise isn't it?
No it's question 9 of review exercise 2.
6. i got limits of pi to pi/3 for the (1+cos)^2 eqtn and between pi/3 and 0 for the 3cos eqtn
phil
Thanks for trying to help but I don't see why that could be the case.
The 1+cost curve is 'smaller' for 0<theta<pi/3, as it is all inside the 3cost curve while the 3cost curve is 'smaller' for pi/3<t<pi.
If I swap the limits around I get 5pi/2.
Thanks for any help.
7. Does this help?
Attached Images

8. (Original post by Fermat)
Does this help?
Aaah the upper limit on the second integral should have been pi/2 instead of pi! Thanks that helps a lot.
I didn't think it would make any real difference as the r=3cosθ curve doesn't have any area for 3pi/2>θ>pi/2, where has the extra area come from? Is a point where r<0 considered to be a point on the opposite side of the pole?
9. (Original post by Gaz031)
Aaah the upper limit on the second integral should have been pi/2 instead of pi! Thanks that helps a lot.
I didn't think it would make any real difference as the r=3cosθ curve doesn't have any area for 3pi/2>θ>pi/2, where has the extra area come from? Is a point where r<0 considered to be a point on the opposite side of the pole?
Taking the limits from 0 -> pi/2 does the upper half-circle.
The lower half-circle comes from pi/2 -> pi.

It looks like your supposition is correct: -ve r comes out on the other side of the pole!

Edit: I got those curves using graphmatica with limits.
Attached Images

10. i'm lost - where did pi/2 come from?
11. (Original post by Phil23)
i'm lost - where did pi/2 come from?
mmm.

If you plot r = [email protected] with values of @ from 0 to pi/2 then you'll get the half-circle shown in the attachment above.

Is that what you meant?

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