Respuesta :
Answer:
The answer is "[tex]8\pi[/tex]"
Step-by-step explanation:
[tex]\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)[/tex]
[tex]\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| \\\\[/tex]
[tex]= (4 \cos v, -4 \sin v, 0)[/tex]
[tex]|r_v \times r_u| = \sqrt{16 \cos^2 v +16 \sin^2 v}\\\\[/tex]
[tex]= \sqrt{16( \cos^2 v + \sin^2 v)}\\\\= \sqrt{16(1)}\\\\= \sqrt{4^2}\\\\=4[/tex]
Calculating the surface area:
[tex]=\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[/tex]
This question is based on the parametrization. Therefore, 8[tex]\pi[/tex] is the surface area as a double integral by using parametrization.
Given:
The portion of the cylinder [tex]x^{2} +y^{2} = 16[/tex] between the planes z=4 and z=5.
Let u=z and v=θ and use cylindrical coordinates to parameterize the surface.
We need to express the area of the surface as a double integral.by using parametrization.
According to the question,
It is given that u=z and v=[tex]\theta[/tex],
Therefore, r(v,u) = (4 Cos v,4 Sin v, u)
Range of v : 0 [tex]\leq[/tex] v [tex]\leq[/tex] 2[tex]\pi[/tex]
Range of u : 4 [tex]\leq[/tex] u [tex]\leq[/tex] 5
[tex]\rightarrow r_v= (-4 \;sin\; v,4\;cos\; v,0)\\\rightarrow r_u=(0,0,1)[/tex]
[tex]\rightarrow r_u \times r_v =\begin{bmatrix} \hat{i}& \hat{j} & \hat{k}\\ -4\; sin\;v & 4\; cos\; v & 0 \\ 0 & 0 & 1\end{bmatrix}\\\\=(4 cos v,\;-4 sin v ,0)\\\\\left | r_v \times r_u \right | = \sqrt{16cos^{2}v+16sin^2v } \\\\= \sqrt{16(cos^{2}v+sin^2v) }\\\\=\sqrt{16}\\\\=4[/tex]
Now calculating the surface area,
[tex]=\int\limits^{2\pi}_0\int\limits^5_4 4 du\; dv\\\\=\int\limits^{2\pi}_0 4[u]^5_4 dv\\\\=\int\limits^{2\pi}_0 4[5-4] dv\\\\\\=\int\limits^{2\pi}_0 4dv\\\\\\=4[v]^{2\pi}_0 = 4(2\pi-0)= 8\pi[/tex]
Therefore, 8[tex]\pi[/tex] is the surface area as a double integral by using parametrization.
For more details, please refer this link:
https://brainly.com/question/15585522
