Question

1. Compute the divergence and curl of each of the following vector fields:
a) hx
2 + y
2
, x2 − y
2
, z2
i
b) hx + y, x − y, zi
2. For any two vector fields F, G show that:
∇.(F × G) = G.(∇ × F) − F.(∇ × G).
3. Given scalar functions u(x, y, z), v(x, y, z), w(x, y, z), φ(x, y, z) and a vector field F(x, y, z) =
hu(x, y, z), v(x, y, z), w(x, y, z)i, prove that
div(φF) = φ(div F) + (∇φ) · F.
4. Find a function f such that F = ∇f where F = hyz, xz, xy + 2zi
5. Let r = hx, y, zi and r = |r|, verify that
a) ∇ · r = 3
b) div (r
n
r) = (n + 3)r
n
6. Given u = xy2
z
2 and v = yz − 3x
2
, find ∇ · [(∇u) × (∇v)]
7. Given the vector field F = he
x
sin y, ex
cos y, zi, show that F is irrotational, hence or
otherwise, find the functions f such that ∇f = F.

Answer 1

Answer on Question #72164 – Math – Calculus

1. Compute the divergence and curl of each of the following vector fields:

a)

$$F = (x ^ {2} + y ^ {2}, x ^ {2} - y ^ {2}, z ^ {2})$$

Solution

$$d i v F = \frac {\partial F _ {x}}{\partial x} + \frac {\partial F _ {y}}{\partial y} + \frac {\partial F _ {z}}{\partial z} = 2 x - 2 y + 2 z$$ $$c u r l F = \left| \begin{array}{c c c} \bar {i} & \bar {j} & \bar {k} \\ \frac {\partial}{\partial x} & \frac {\partial}{\partial y} & \frac {\partial}{\partial z} \\ F _ {x} & F _ {y} & F _ {z} \end{array} \right| = (2 x + 2 y) \bar {k}$$

b)

$$F = (x + y, x - y, z)$$

Solution

$$d i v F = \frac {\partial F _ {x}}{\partial x} + \frac {\partial F _ {y}}{\partial y} + \frac {\partial F _ {z}}{\partial z} = 1 - 1 + 1 = 1$$ $$c u r l F = \left| \begin{array}{c c c} \bar {i} & \bar {j} & \bar {k} \\ \frac {\partial}{\partial x} & \frac {\partial}{\partial y} & \frac {\partial}{\partial z} \\ F _ {x} & F _ {y} & F _ {z} \end{array} \right| = 0$$

2. For any two vector fields $F, G$ show that:

$$\nabla \cdot (F \times G) = G \cdot (\nabla \times F) - F \cdot (\nabla \times G)$$

Solution

$$F \times G = \left| \begin{array}{c c c} \bar {i} & \bar {j} & \bar {k} \\ F _ {x} & F _ {y} & F _ {z} \\ G _ {x} & G _ {y} & G _ {z} \end{array} \right| = \big (F _ {y} \cdot G _ {z} - F _ {z} \cdot G _ {y} \big) \bar {i} - (F _ {x} \cdot G _ {z} - F _ {z} \cdot G _ {x}) \bar {j} + \big (F _ {x} \cdot G _ {y} - F _ {y} \cdot G _ {x} \big) \bar {k}$$

Answer on Question #72164 – Math – Calculus

$$\nabla \cdot (F \times G) = \frac {\partial (F _ {y} \cdot G _ {z} - F _ {z} \cdot G _ {y})}{\partial x} - \frac {\partial (F _ {x} \cdot G _ {z} - F _ {z} \cdot G _ {x})}{\partial y} + \frac {\partial (F _ {x} \cdot G _ {y} - F _ {y} \cdot G _ {x})}{\partial z} =$$ $$= G _ {z} \cdot \frac {\partial F _ {y}}{\partial x} + F _ {y} \cdot \frac {\partial G _ {z}}{\partial x} - G _ {y} \cdot \frac {\partial F _ {z}}{\partial x} - F _ {z} \cdot \frac {G _ {y}}{\partial x} - G _ {z} \cdot \frac {\partial F _ {x}}{\partial y} - F _ {x} \cdot \frac {\partial G _ {z}}{\partial y} + G _ {x} \cdot \frac {\partial F _ {z}}{\partial y} + F _ {z} \cdot \frac {\partial G _ {x}}{\partial y} +$$ $$+ G _ {y} \cdot \frac {\partial F _ {x}}{\partial z} + F _ {x} \cdot \frac {\partial G _ {y}}{\partial z} - G _ {x} \cdot \frac {\partial F _ {y}}{\partial z} - F _ {y} \cdot \frac {\partial G _ {x}}{\partial z} =$$ $$= G _ {x} \cdot \left(\frac {\partial F _ {z}}{\partial y} - \frac {\partial F _ {y}}{\partial z}\right) - G _ {y} \cdot \left(\frac {\partial F _ {z}}{\partial x} - \frac {\partial F _ {x}}{\partial z}\right) + G _ {z} \cdot \left(\frac {\partial F _ {y}}{\partial x} - \frac {\partial F _ {x}}{\partial y}\right) - F _ {x} \cdot \left(\frac {\partial G _ {z}}{\partial y} - \frac {\partial G _ {y}}{\partial z}\right) +$$ $$+ F _ {y} \cdot \left(\frac {\partial G _ {z}}{\partial x} - \frac {\partial G _ {x}}{\partial z}\right) - F _ {z} \cdot \left(\frac {\partial G _ {y}}{\partial x} - \frac {\partial G _ {x}}{\partial y}\right)$$ $$\nabla \times F = \left| \begin{array}{c c c} \bar {\iota} & \bar {\jmath} & \bar {k} \\ \frac {\partial}{\partial x} & \frac {\partial}{\partial y} & \frac {\partial}{\partial z} \\ F _ {x} & F _ {y} & F _ {z} \end{array} \right| = \left(\frac {\partial F _ {z}}{\partial y} - \frac {\partial F _ {y}}{\partial z}\right) \bar {\iota} - \left(\frac {\partial F _ {z}}{\partial x} - \frac {\partial F _ {x}}{\partial z}\right) \bar {\jmath} + \left(\frac {\partial F _ {y}}{\partial x} - \frac {\partial F _ {x}}{\partial y}\right) \bar {k}$$ $$G \cdot (\nabla \times F) = G _ {x} \cdot \left(\frac {\partial F _ {z}}{\partial y} - \frac {\partial F _ {y}}{\partial z}\right) - G _ {y} \cdot \left(\frac {\partial F _ {z}}{\partial x} - \frac {\partial F _ {x}}{\partial z}\right) + G _ {z} \cdot \left(\frac {\partial F _ {y}}{\partial x} - \frac {\partial F _ {x}}{\partial y}\right)$$

Analogically,

$$F \cdot (\nabla \times G) = F _ {x} \cdot \left(\frac {\partial G _ {z}}{\partial y} - \frac {\partial G _ {y}}{\partial z}\right) - F _ {y} \cdot \left(\frac {\partial G _ {z}}{\partial x} - \frac {\partial G _ {x}}{\partial z}\right) + F _ {z} \cdot \left(\frac {\partial G _ {y}}{\partial x} - \frac {\partial G _ {x}}{\partial y}\right)$$

Finally,

$$G \cdot (\nabla \times F) - F \cdot (\nabla \times G) = G _ {x} \cdot \left(\frac {\partial F _ {z}}{\partial y} - \frac {\partial F _ {y}}{\partial z}\right) - G _ {y} \cdot \left(\frac {\partial F _ {z}}{\partial x} - \frac {\partial F _ {x}}{\partial z}\right) + G _ {z} \cdot \left(\frac {\partial F _ {y}}{\partial x} - \frac {\partial F _ {x}}{\partial y}\right) -$$ $$- F _ {x} \cdot \left(\frac {\partial G _ {z}}{\partial y} - \frac {\partial G _ {y}}{\partial z}\right) + F _ {y} \cdot \left(\frac {\partial G _ {z}}{\partial x} - \frac {\partial G _ {x}}{\partial z}\right) - F _ {z} \cdot \left(\frac {\partial G _ {y}}{\partial x} - \frac {\partial G _ {x}}{\partial y}\right) = \nabla \cdot (F \times G)$$

Answer on Question #72164 – Math – Calculus

3. Given scalar functions $u(x,y,z), v(x,y,z), w(x,y,z), \varphi(x,y,z)$ and a vector field

$$F(x, y, z) = u(x, y, z), v(x, y, z), w(x, y, z),$$

prove that

$$\operatorname{div}(\varphi F) = \varphi(\operatorname{div} F) + (\nabla \varphi) \cdot F$$

Solution

$$\operatorname{div} F = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z}$$ $$\varphi(\operatorname{div} F) = \varphi\left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z}\right)$$ $$\nabla \varphi = \frac{\partial \varphi}{\partial x} \bar{i} + \frac{\partial \varphi}{\partial y} \bar{j} + \frac{\partial \varphi}{\partial z} \bar{k}$$ $$(\nabla \varphi) \cdot F = \frac{\partial \varphi}{\partial x} \cdot u + \frac{\partial \varphi}{\partial y} \cdot v + \frac{\partial \varphi}{\partial z} \cdot w$$ $$\varphi(\operatorname{div} F) + (\nabla \varphi) \cdot F = \varphi\left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z}\right) + \frac{\partial \varphi}{\partial x} \cdot u + \frac{\partial \varphi}{\partial y} \cdot v + \frac{\partial \varphi}{\partial z} \cdot w$$ $$\operatorname{div}(\varphi F) = \operatorname{div}(\varphi u + \varphi v + \varphi w) = \frac{\partial (\varphi u)}{\partial x} + \frac{\partial (\varphi v)}{\partial y} + \frac{\partial (\varphi w)}{\partial z} =$$ $$= \frac{\partial \varphi}{\partial x} \cdot u + \frac{\partial \varphi}{\partial y} \cdot v + \frac{\partial \varphi}{\partial z} \cdot w + \frac{\partial u}{\partial x} \cdot \varphi + \frac{\partial v}{\partial y} \cdot \varphi + \frac{\partial w}{\partial z} \cdot \varphi =$$ $$= \varphi(\operatorname{div} F) + (\nabla \varphi) \cdot F$$

4. Find a function $f$ such that $F = \nabla f$, where $F = (yz, xz, xy + 2z)$

Solution

$$\nabla f = \frac{\partial f}{\partial x} \bar{i} + \frac{\partial f}{\partial y} \bar{j} + \frac{\partial f}{\partial z} \bar{k} = F$$ $$\frac{\partial f}{\partial x} = yz; \frac{\partial f}{\partial y} = xz; \frac{\partial f}{\partial z} = xy + 2z$$

Answer on Question #72164 – Math – Calculus

Answer:

$$f = (xyz, xyz, xyz + z^2)$$

5. Let $r = (x, y, z)$ and $r = |r|$, verify that

a)

$$\nabla \cdot r = 3$$

Solution

$$\nabla \cdot r = \frac{\partial r_x}{\partial x} + \frac{\partial r_y}{\partial y} + \frac{\partial r_z}{\partial z} = 1 + 1 + 1 = 3$$

b)

$$div(r^n \cdot r) = (n + 3)r^n$$

Solution

$$\begin{aligned} r^n \cdot r &= \left(\sqrt{x^2 + y^2 + z^2}\right)^n \cdot (x, y, z) = \left[ x \left(\sqrt{x^2 + y^2 + z^2}\right)^n, y \left(\sqrt{x^2 + y^2 + z^2}\right)^n, z \left(\sqrt{x^2 + y^2 + z^2}\right)^n \right] \\ div(r^n \cdot r) &= \left(\sqrt{x^2 + y^2 + z^2}\right)^n + x \cdot 2x \cdot \frac{n}{2}(x^2 + y^2 + z^2)^{\frac{n}{2} - 1} + \\ &\quad + \left(\sqrt{x^2 + y^2 + z^2}\right)^n + y \cdot 2y \cdot \frac{n}{2}(x^2 + y^2 + z^2)^{\frac{n}{2} - 1} + \\ &\quad + \left(\sqrt{x^2 + y^2 + z^2}\right)^n + z \cdot 2z \cdot \frac{n}{2}(x^2 + y^2 + z^2)^{\frac{n}{2} - 1} = \\ &= 3r^n + nr^{2\left(\frac{n}{2} - 1\right)}(x^2 + y^2 + z^2) = 3r^n + nr^{n-2}r^2 = 3r^n + nr^n = (n + 3)r^n \end{aligned}$$

6. Given $u = xy^2z^2$ and $v = yz - 3x^2$, find $\nabla \cdot [(\nabla u) \times (\nabla v)]$

Solution

$$\nabla u = \frac{\partial u}{\partial x} \bar{i} + \frac{\partial u}{\partial y} \bar{j} + \frac{\partial u}{\partial z} \bar{k} = (y^2z^2) \bar{i} + (2xyz^2) \bar{j} + (2xy^2z) \bar{k}$$

Answer on Question #72164 – Math – Calculus

$$\nabla v = \frac{\partial v}{\partial x} \bar{i} + \frac{\partial v}{\partial y} \bar{j} + \frac{\partial v}{\partial z} \bar{k} = (-6x) \bar{i} + (z) \bar{j} + (y) \bar{k}$$ $$(\nabla u) \times (\nabla v) = \left| \begin{array}{ccc} \bar{i} & \bar{j} & \bar{k} \\ (\nabla u)_x & (\nabla u)_y & (\nabla u)_z \\ (\nabla v)_x & (\nabla v)_y & (\nabla v)_z \end{array} \right| = (2xyz^2 \cdot y - z \cdot 2xy^2z) \bar{i} - (y^2z^2 \cdot y + 2xy^2z \cdot 6x) \bar{j} + (y^2z^2 \cdot z + 2xyz^2 \cdot 6x) \bar{k} = -(y^3z^2 + 12x^2y^2z) \bar{j} + (y^2z^3 + 12x^2yz^2) \bar{k}$$ $$\nabla \cdot [(\nabla u) \times (\nabla v)] = \frac{\partial [(\nabla u) \times (\nabla v)]_x}{\partial x} + \frac{\partial [(\nabla u) \times (\nabla v)]_y}{\partial y} + \frac{\partial [(\nabla u) \times (\nabla v)]_z}{\partial z}$$

Answer:

$$\nabla \cdot [(\nabla u) \times (\nabla v)] = -3y^2z^2 - 24x^2yz + 3y^2z^2 + 24x^2yz = 0$$

7. Given the vector field $F = (e^x \sin y, e^x \cos y, z)$, show that $F$ is irrotational, hence or otherwise, find the functions $f$ such that $\nabla f = F$

Solution

$$\begin{array}{l} \text{curl } F = \left| \begin{array}{ccc} \bar{i} & \bar{j} & \bar{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z \end{array} \right| = \left| \begin{array}{ccc} \bar{i} & \bar{j} & \bar{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ e^x \sin y & e^x \cos y & z \end{array} \right| = \\ = (e^x \cos y - e^x \cos y) \bar{k} = 0 \\ \nabla f = \frac{\partial f}{\partial x} \bar{i} + \frac{\partial f}{\partial y} \bar{j} + \frac{\partial f}{\partial z} \bar{k} = F \\ \frac{\partial f}{\partial x} = e^x \sin y; \frac{\partial f}{\partial y} = e^x \cos y; \frac{\partial f}{\partial z} = z \\ \end{array}$$

Answer:

$$f = \left(e^x \sin y, e^x \sin y, \frac{z^2}{2}\right)$$

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