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Chapter 14: Q. 15 (page 1132)

Use the same vector field as in Exercise 13, and compute the k-component of the curl of F(x, y).

Short Answer

Expert verified

$The k - component of the curl of F (x, y) is - sinx siny - x^{2} e^{y} .$

Step by step solution

01

Step 1. Given information is

$F (x, y) = x^{2} e^{y} i + \cos (x) \sin (y) j$

02

Step 2. Curl of F

$The curl of a vector field F (x, y) = F_{1} (x, y) i + F_{2} (x, y) j in ℝ^{2} is defined as follows : curl F = (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) k Hence the k component of the curl of F (x, y) is, \frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y} For the vector field, F (x, y) = x^{2} e^{y} i + \cos (x) \sin (y) j F_{1} (x, y) = x^{2} e^{y} F_{2} (x, y) = \cos (x) \sin (y) Now, first find \frac{\partial F_{2}}{\partial x} and \frac{\partial F_{1}}{\partial y} Then, \frac{\partial F_{2}}{\partial x} = \frac{\partial (\cos x \sin y)}{\partial x} = - sinx siny and, \frac{\partial F_{1}}{\partial y} = \frac{\partial (x^{2} e^{y})}{\partial y} = x^{2} e^{y} Then, \frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y} = - sinx siny - x^{2} e^{y} Therefore, the k - component of the curl of F (x, y) is - sinx siny - x^{2} e^{y} .$

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Most popular questions from this chapter

Find $\int_{S} 1 d S$ , where S is the portion of the surface with equation $x = e^{y z} - e^{- y z}$ that lies on the positive side of the circle of radius 3 and centered at the origin in the yz-plane.

$F (x, y, z) = i + j + k$ , where S is the lower half of the unit sphere, with n pointing outwards.

What is the difference between the graphs of

$G (x, y, z) = 2 i - 3 j + z k a n d F (x, y, z) = - 2 i + 3 j - z k$

What are the inputs of a vector field in $ℝ^{3}$ ?

Write two different normal vectors for a smooth surface S given by (x, y, g(x, y)) at the point $(x_{0}, y_{0}, g (x_{0}, y_{0})) .$

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