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Chapter 14: Q. 28 (page 1132)

Compute the curl of the vector fields:
$G (x, y, z) = (2 y + 3 z) i + 2 x y^{2} j + (z x - y) k$ .

Short Answer

Expert verified

The curl is,

$- i + (3 - z) j + (2 y^{2} - 2) k$ .

Step by step solution

01

Step 1. Given Information.

The vector field is,

$G (x, y, z) = (2 y + 3 z) i + 2 x y^{2} j + (z x - y) k$ .

02

Step 2. Finding the curl for the vector field.

Consider the vector field,

$G (x, y, z) = (2 y + 3 z) i + 2 x y^{2} j + (z x - y) k$

If $G (x, y, z) = g_{1} (x, y, z) i + g_{2} (x, y, z) j + g_{3} (x, y, z)$ , then,

curl $G = 鈭� 脳 G$

$= |\begin{matrix} i & j & k \\ \frac{鈭�}{鈭� x} & \frac{鈭�}{鈭� y} & \frac{鈭�}{鈭� z} \\ g_{1} (x, y, z) & g_{2} (x, y, z) & g_{3} (x, y, z) \end{matrix}| = |\begin{matrix} i & j & k \\ \frac{鈭�}{鈭� x} & \frac{鈭�}{鈭� y} & \frac{鈭�}{鈭� z} \\ (2 y + 3 z) & 2 x y^{2} & (z x - y) \end{matrix}| = i (\frac{鈭�}{鈭� y} (z x - y) - \frac{鈭�}{鈭� z} (2 x y^{2})) + j (\frac{鈭�}{鈭� z} (2 y + 3 z) - \frac{鈭�}{鈭� x} (z x - y)) + k (\frac{鈭�}{鈭� x} (2 x y^{2}) - \frac{鈭�}{鈭� y} (2 y + 3 z)) = ((- 1) - 0) i + (3 - z) j + (2 y^{2} - 2) k = - i + (3 - z) j + (2 y^{2} - 2) k$

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Most popular questions from this chapter

S is the portion of the saddle surface determined by z = x² 鈭� y² that lies above and/or below the annulus in the xy-plane determined by the circles with radii

$\frac{\sqrt{3}}{2} and \sqrt{2}$

and centered at the origin.

$Compute the divergence of the vector fields in Exercises 17 鈥� 22 . G (x, y, z) = (x^{2} + y 鈭� z) i 鈭� 2 y \cos zj + e^{x^{2} + y^{2} + z^{2}} k$

What is the difference between the graphs of

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

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

What is the difference between the graphs of

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

See all solutions

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