Question

number 11

Answer 1

@Zepdrix

Answer 2

Yo, im not smart enough for this xD

Answer 3

Did you get help?

Answer 4

Clearly we can just say that \(S = 2 \pi r \cdot h \).

But, obviously, this is an exercise in playing around with parameterised surfaces and we have:

\(\mathbf r(u,v) = <2\cos u , 2\sin u , v> \qquad u \in [0, 2 \pi] \qquad v \in [0,4] \)

[Aside: this is an undercover cylindrical co-ordinate parameterisation ;) Think \(2 \leftrightarrow r, u \leftrightarrow \theta, v \leftrightarrow z\)]

But getting on with it, the tangent vector with \(v\) held constant is: \(\delta \mathbf r_u =\frac{\partial \mathbf r}{\partial u} ~ \delta u = <- 2 \sin u, 2 \cos u , 0> \delta u\)

Likewise with \(u\) held constant: \(\delta \mathbf r_v =\frac{\partial \mathbf r}{\partial v} ~ \delta v = <0, 0 , 1> \delta v\)

So infinitessimal area vector is the vector product, which is uusally done as a matrix determinant:

\(\delta \mathbf A = \delta \mathbf r_u \times \delta \mathbf r_v = 2 <\cos u , \sin u , 0> \delta u \delta v\)

We're interested in the absolute value of the area \(\delta A\), irrespective of orientation:

\(\delta A = \delta | \mathbf A| = | 2 <\cos u , \sin u , 0> | ~ \delta u ~ \delta v = 2 ~ \delta u ~ \delta v\)


So:

\(A = \iint_A dA = 2 \int\limits_{u = 0}^{2 \pi} ~ \left( \int\limits_{v = 0}^{4} ~ dv \right) ~ du \)

\(= 2 \cdot 2 \pi \cdot 4 = 16 \pi \)