 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-1/section-b/7/ii/solution

1 < z < 2

0 < r < 2 - z^{2}

and

0 < θ < 2 π

k z

r d r d θ d z

. Therefore

M = \int_{1}^{2} \int_{0}^{2 π} \int_{0}^{2 - z^{2}} k zr d r d θ d z = πk \int_{1}^{2} z (2 - z^{2}) d z = \frac{πk}{4} .

R = 2

and

c = 1

For its curved surface area, write

z = 2 - r^{2}

with

0 \leq r \leq 1

. The surface integral of one becomes

A = \int_{0}^{2 π} \int_{0}^{1} 1 + (\frac{d z}{d r})^{2} r d r d θ = 2 π 2 \int_{0}^{1} \frac{r}{2 - r ^{2}} d r = 2 π (2 - 2) .

2 π R h

, for

h = 2 - 1

. The planar closing disk is not part of the requested curved surface.

 New to topics? Read the docs here!