URGENT: calculus

Hi,

I am completely lost on question 2 and 4 on this sheet.

http://www.maths.ox.ac.uk/~kirchhei/2007_4.pdf

I have tried everything I can think of over the past 4 days and I am getting nowhere. Please help and if you could explain what you are doing as you go along so that I can hopefully understand the process that would be great.

Thanks

Reply 1

gyrase

3

Qn2.

Let the density of the cone be p.

Height of cone = a cot(alpha)
Radius of cone at height z (where 0 <= z <= a cot(alpha)) = z*tan(alpha)

Imagine splitting the cone up into discs parallel to the base, then the discs into annuli.

M
=

\int_0^{a \cot\alpha} \int_0^{z\tan{\alpha}}

2pi r p dr dz
= (1/3)a^3 p pi cot(alpha)

Gravitational field magnitude
=

\int_0^{a \cot\alpha} \int_0^{z\tan{\alpha}}

2pi r p [G/(z^2 + r^2)] [z/sqrt(z^2 + r^2)] dr dz
= 2a G p pi cos(alpha) tan(alpha/2)
= (12GM/a^2) sin^2(alpha/2)

Reply 2

xxrachxx

OP

Decota

Qn2.

Gravitational field magnitude
=

\int_0^{a \cot\alpha} \int_0^{z\tan{\alpha}}

2pi r p [G/(z^2 + r^2)] [z/sqrt(z^2 + r^2)] dr dz
= 2a G p pi cos(alpha) tan(alpha/2)
= (12GM/a^2) sin^2(alpha/2)

what identity did you use to get the tan(alpha/2)?

Reply 3

gyrase

3

xxrachxx

what identity did you use to get the tan(alpha/2)?

When you evaluate the integral obtaining;
cot(t)-cot(t)cos(t), (t = alpha for convenience), then factor cos(t) out and remaining cosec(t)-cot(t) = tan(t/2)

edit; my method outlined above is a bit vague and long-winded. Not the most elegant, maybe someone else knows of a better way.

Reply 4

xxrachxx

OP

Decota

When you evaluate the integral obtaining;
cot(t)-cot(t)cos(t), (t = alpha for convenience), then factor cos(t) out and remaining cosec(t)-cot(t) = tan(t/2)

edit; my method outlined above is a bit vague and long-winded. Not the most elegant, maybe someone else knows of a better way.

No it's ok I get now

thanks

URGENT: calculus

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