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**zizou1089** Find the surface area of that part of the cone z^2 = x^2 + y^2 which lies inside the cylinder x^2 + y^2 = 2x

OK SO: z is a circle (x − 1)^2 + y^2 = 1. We use polar coordinates for x

and y

x = pcos(t) , y = psin(t) .

Then, since we are interested in a surface living on the cone, its equation yields z^2 = p^2.

The sought surface is made of two parts, one for positive z i.e. z = p and one for negative z i.e. z = −p. We need to determine the bounds for p and t. This is given by the fact that we are interested in the interior of the cylinder

x^2 + y^2 − 2x ≤ 0. This gives

2 − 2pcos(t) ≤ 0

Combined with the fact that p ≥ 0 in polar coordinates we get

0 ≤ p ≤ 2 cos (t)

which also shows that 0 ≤ pcos(t) so t E [−pi/2,pi/2 ]. Collecting everything the parametrization for the top part is

**−>**

**F 1 : [0, 2 cos (t)] × [−pi/2,pi/2] → R^3**

**(p,(t)) → pcos(t)i+psin(t)j+pk**

**and for the bottom part **

**−>**

**F 1 : [0, 2 cos (t)] × [−pi/2,pi/2] → R^3**

**(p,(t)) → pcos(t)i+psin(t)j-pk**

I JUST DONT UNDERSTAND HOW THE PARAMETRIZATION OF THE TOP AND BOTTOM PART HAVE BEEN WORKED OUT!