Answered: Surface integrals using a parametric…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

Surface integrals using a parametric description Evaluate the surface integral ∫∫_S ƒ dS using a parametric description of the surface.

ƒ(x, y, z) = x² + y², where S is the hemisphere x² + y² + z² = 36, for z ≥ 0

Expert Solution

Step 1

Given: $f (x, y, z) = x^{2} + y^{2}; S : x^{2} + y^{2} + z^{2} = 36, z \geq 0$
To find: $\iint_{s} f (x, y, z) d S$ using a parametric description of the surface.

Step 2

We have,
$f (x, y, z) = x^{2} + y^{2}; S : x^{2} + y^{2} + z^{2} = 36, z \geq 0$
Use the spherical coordinate with $x = ρ \sin ϕ \cos θ, y = ρ \sin ϕ \sin θ and z = ρ \cos ϕ$ .

Parametrizing the given curve, we get
$r (u, v) = <6 \sin u \cos v, 6 \sin u \sin v, 6 \cos u> where 0 \leq u \leq \frac{π}{2}, 0 \leq v \leq 2 π Now, r_{u} = <6 \cos u \cos v, 6 \cos u \sin v, - 6 \sin u> r_{v} = <- 6 \sin u \sin v, 6 \sin u \cos v, 0> ∴ r_{u} \times r_{v} = |\begin{matrix} i & j & k \\ 6 \cos u \cos v & 6 \cos u \sin v & - 6 \sin u \\ - 6 \sin u \sin v & 6 \sin u \cos v & 0 \end{matrix}| = i [36 \sin^{2} u \cos v] - j [- 36 \sin^{2} u \sin v] + k [36 \sin u \cos u \cos^{2} v + 36 \sin u \cos u \sin^{2} v] = i [36 \sin^{2} u \cos v] + j [36 \sin^{2} u \sin v] + k [36 \sin u \cos u (\sin^{2} v + \cos^{2} v)] = i [36 \sin^{2} u \cos v] + j [36 \sin^{2} u \sin v] + k [36 \sin u \cos u]$

Step by step

Solved in 3 steps

Knowledge Booster

$Triple Integral$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions