Q. 29 Integrate the given function ove... [FREE SOLUTION]

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Chapter 14: Q. 29 (page 1119)

Integrate the given function over the accompanying surface in Exercises 27鈥�34.
$f (x, y, z) = x y z^{2}$ , where S is the portion of the cone $\sqrt{3} z = \sqrt{x^{2} + y^{2}}$ that lies within the sphere of radius 4 and centered at the origin.

Short Answer

Expert verified

The integration of the function $f (x, y, z) = x y z^{2}$ is $0$ .

Step by step solution

Step 1. Given Information.

The function:

$f (x, y, z) = x y z^{2} \sqrt{3} z = \sqrt{x^{2} + y^{2}}$

Step 2. Formula for finding the integral.

The formula for finding the integral over the smooth surface is given by,

${鈭�}_{S} d S = 鈭� {鈭�}_{D} f (x (u, v), y (u, v), z (u, v)) ||r_{u} 脳 r_{v}|| d u d v$

Step 3. Find ru,rv

The surface is parametrized as follows:

$r (u, v) = (u \cos v, u \sin v, \frac{u}{\sqrt{3}}) r_{u} = \frac{鈭� r}{鈭� u} = (\cos v, \sin v, \frac{1}{\sqrt{3}}) r_{v} = \frac{鈭� r}{鈭� v} = (- u \sin v, u \cos v, 0)$

Step 4. Find ru×rv

$r_{u} 脳 r_{v} = (\cos v, \sin v, \frac{1}{\sqrt{3}}) 脳 (- u \sin v, u \cos v, 0) = - \frac{1}{\sqrt{3}} u \cos v i - \frac{1}{\sqrt{3}} u \sin v j + u k ||r_{u} 脳 r_{v}|| = \sqrt{{(- \frac{1}{\sqrt{3}} u \cos v)}^{2} + (- \frac{1}{\sqrt{3}} u \sin v) + u^{2}} = \sqrt{\frac{1}{3} u^{2} + u^{2}} = \sqrt{\frac{4 u^{2}}{3}} = \frac{2 u}{\sqrt{3}}$

Step 5. Find the parametrization function.

$f (x, y, z) = x y z^{2} f (x (u, v), y (u, v), z (u, v)) = (u \cos v) (u \sin v) {(\frac{u}{\sqrt{3}})}^{2} = \frac{u^{4}}{3} \sin v \cos v$

Step 6. Find the region of integration and integrate.

The region of integration is given by,

$D = {(u, v) 0 鈮� u 鈮� 4, 0 鈮� v 鈮� 2 蟺} {鈭�}_{S} d S = 鈭� {鈭�}_{D} f (x (u, v), y (u, v), z (u, v)) ||r_{u} 脳 r_{v}|| d u d v = {鈭�}_{0}^{2 蟺} {鈭�}_{0}^{4} ({\frac{u}{3}}^{4} \sin v \cos v (\frac{2 u}{\sqrt{3}})) d u d v = {鈭�}_{0}^{2 蟺} \frac{2}{3 \sqrt{3}} \sin v \cos v {[\frac{u^{6}}{6}]}_{0}^{4} d v = {鈭�}_{0}^{2 蟺} \frac{2}{3 \sqrt{3}} \sin v \cos v (\frac{2048}{3}) d v = \frac{4096}{9 \sqrt{3}} {鈭�}_{0}^{2 蟺} \sin v \cos v d v = \frac{4096}{9 \sqrt{3}} {[\frac{1}{2} \sin^{2} v]}_{0}^{2 蟺} = \frac{4096}{9 \sqrt{3}} (0) = 0$

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91影视

Integrate the given function over the accompanying surface in Exercises 27鈥�34.
$f (x, y, z) = x y z^{2}$ , where S is the portion of the cone $\sqrt{3} z = \sqrt{x^{2} + y^{2}}$ that lies within the sphere of radius 4 and centered at the origin.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Formula for finding the integral.

Step 3. Find ru,rv

Step 4. Find ru×rv

Step 5. Find the parametrization function.

Step 6. Find the region of integration and integrate.

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Integrate the given function over the accompanying surface in Exercises 27鈥�34.f(x,y,z)=xyz2, where S is the portion of the cone 3z=x2+y2 that lies within the sphere of radius 4 and centered at the origin.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Formula for finding the integral.

Step 3. Find ru,rv

Step 4. Find ru×rv

Step 5. Find the parametrization function.

Step 6. Find the region of integration and integrate.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Geometry

Discrete Mathematics

Decision Maths

Logic and Functions

Calculus

Study anywhere. Anytime. Across all devices.

Integrate the given function over the accompanying surface in Exercises 27鈥�34.
$f (x, y, z) = x y z^{2}$ , where S is the portion of the cone $\sqrt{3} z = \sqrt{x^{2} + y^{2}}$ that lies within the sphere of radius 4 and centered at the origin.