Question

Use a parametrization to express the area of the surface as a double integral. Then evaluate the integral. The portion of the cylinder x^2+y^2=16 between the planes z=4 and z=5
Let u=z and v=θ and use cylindrical coordinates to parameterize the surface. Set up the double integral to find the surface area.

Answer 1

Answer:

The answer is "$8\pi$"

Step-by-step explanation:

$\to r(v,u) =(4 \cos v, 4 \sin v,u) \\\\0\leq v \leq 2\pi, \ \ 4\leq u\leq 5\\\\\to r_v= (-4 \sin v, 4 \cos v, 0)\\\\\to r_u=(0,0,1)$

$\to r_v\times r_u = \left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\-4 \sin v& 4 \cos v& 0\\0&0&1\end{array}\right|$

                $= (4 \cos v, -4 \sin v, 0)$

$|r_v \times r_u| = \sqrt{16 \cos^2 v +16 \sin^2 v}$

             $= \sqrt{16( \cos^2 v + \sin^2 v)}\\\\= \sqrt{16(1)}\\\\= \sqrt{4^2}\\\\=4$

Calculating the surface area:

$=\int^{2\pi}_{0} \int^{5}_{4} 4 \ du \ dv \\\\=\int^{2\pi}_{0} 4[4]^{5}_{4} \ dv \\\\=\int^{2\pi}_{0} 4[5-4] \ dv \\\\=\int^{2\pi}_{0} 4 \ dv \\\\=4 [v]^{2\pi}_{0}\\\\= 4 \times 2\pi\\\\= 8\pi$