/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Explain how to compute the curl ... [FREE SOLUTION] | 91Ó°ÊÓ

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Explain how to compute the curl of the vector field \(\mathbf{F}=\langle f, g, h\rangle\).

Short Answer

Expert verified
Given the vector field \(\mathbf{F}=\langle x^2 + 3y, xy - z^2, 2x + y^3\rangle\), compute the curl of the vector field. Solution: Step 1: Write down the given vector field and the curl formula The given vector field is \(\mathbf{F}=\langle x^2 + 3y, xy - z^2, 2x + y^3\rangle\). The curl of the vector field is given by the formula: $$ \text{curl } \mathbf{F} = \nabla \times \mathbf{F} = \left\langle\frac{\partial h}{\partial y} - \frac{\partial g}{\partial z}, \frac{\partial f}{\partial z} - \frac{\partial h}{\partial x}, \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}\right\rangle $$ Step 2: Compute the partial derivatives 1. \(\frac{\partial f}{\partial x} = 2x\), \(\frac{\partial f}{\partial y} = 3\), and \(\frac{\partial f}{\partial z} = 0\) 2. \(\frac{\partial g}{\partial x} = y\), \(\frac{\partial g}{\partial y} = x\), and \(\frac{\partial g}{\partial z} = -2z\) 3. \(\frac{\partial h}{\partial x} = 2\), \(\frac{\partial h}{\partial y} = 3y^2\), and \(\frac{\partial h}{\partial z} = 0\) Step 3: Plug the partial derivatives into the curl formula $$ \text{curl } \mathbf{F} = \left\langle\frac{\partial h}{\partial y} - \frac{\partial g}{\partial z}, \frac{\partial f}{\partial z} - \frac{\partial h}{\partial x}, \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}\right\rangle = \left\langle 3y^2 - (-2z), 0-2, y-3\right\rangle $$ Step 4: Simplify the expression $$ \text{curl } \mathbf{F} = \left\langle 3y^2 + 2z, -2, y-3 \right\rangle $$ The curl of the given vector field \(\mathbf{F}=\langle x^2 + 3y, xy - z^2, 2x + y^3\rangle\) is \(\text{curl } \mathbf{F} = \left\langle 3y^2 + 2z, -2, y-3 \right\rangle\).

Step by step solution

01

Write down the given vector field and the curl formula

We are given the vector field \(\mathbf{F}=\langle f, g, h\rangle\). The curl of the vector field is given by the formula: $$ \text{curl } \mathbf{F} = \nabla \times \mathbf{F} = \left\langle\frac{\partial h}{\partial y} - \frac{\partial g}{\partial z}, \frac{\partial f}{\partial z} - \frac{\partial h}{\partial x}, \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}\right\rangle $$
02

Compute the partial derivatives

Compute the partial derivatives of the components of the vector field: 1. \(\frac{\partial f}{\partial x}\), \(\frac{\partial f}{\partial y}\), and \(\frac{\partial f}{\partial z}\) 2. \(\frac{\partial g}{\partial x}\), \(\frac{\partial g}{\partial y}\), and \(\frac{\partial g}{\partial z}\) 3. \(\frac{\partial h}{\partial x}\), \(\frac{\partial h}{\partial y}\), and \(\frac{\partial h}{\partial z}\)
03

Plug the partial derivatives into the curl formula

Substitute the computed partial derivatives into the curl formula: $$ \text{curl } \mathbf{F} = \left\langle\frac{\partial h}{\partial y} - \frac{\partial g}{\partial z}, \frac{\partial f}{\partial z} - \frac{\partial h}{\partial x}, \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}\right\rangle $$
04

Simplify the expression

Simplify the expression, if possible, to get the curl of the vector field: $$ \text{curl } \mathbf{F} = \left\langle A, B, C \right\rangle $$ The final result will be a vector field that represents the curl of the given vector field \(\mathbf{F}=\langle f, g, h\rangle\).

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Most popular questions from this chapter

Consider the sphere \(x^{2}+y^{2}+z^{2}=4\) and the cylinder \((x-1)^{2}+y^{2}=1,\) for \(z \geq 0\) a. Find the surface area of the cylinder inside the sphere. b. Find the surface area of the sphere inside the cylinder.

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