Answer \( \nabla\cdot\mathbf{F} = 2 x + 2 y + 2 z,\quad \nabla\times\mathbf{F} = \langle 0, 0, 0 \rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the divergence and curl of \( \displaystyle \mathbf{F} = \langle x z^{2}, - y, x y z \rangle \).
\[ \frac{d}{d y} \left(- y\right) + \frac{\partial}{\partial x} x z^{2} + \frac{\partial}{\partial z} x y z = x y + z^{2} - 1 \]
div F.✓ Proved
\[ \left[\begin{matrix}- \frac{d}{d z} \left(- y\right) + \frac{\partial}{\partial y} x y z\\\frac{\partial}{\partial z} x z^{2} - \frac{\partial}{\partial x} x y z\\\frac{d}{d x} \left(- y\right) - \frac{\partial}{\partial y} x z^{2}\end{matrix}\right] = \left[\begin{matrix}x z\\2 x z - y z\\0\end{matrix}\right] \]
curl F, component by component.✓ Proved
Answer \( \nabla\cdot\mathbf{F} = x y + z^{2} - 1,\quad \nabla\times\mathbf{F} = \langle x z, 2 x z - y z, 0 \rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the divergence and curl of \( \displaystyle \mathbf{F} = \langle - 3 e^{x}, - 2 \cos{\left(z \right)}, 3 x y z \rangle \).
\[ \frac{\partial}{\partial z} 3 x y z + \frac{d}{d x} \left(- 3 e^{x}\right) + \frac{d}{d y} \left(- 2 \cos{\left(z \right)}\right) = 3 x y - 3 e^{x} \]
div F.✓ Proved
\[ \left[\begin{matrix}\frac{\partial}{\partial y} 3 x y z - \frac{d}{d z} \left(- 2 \cos{\left(z \right)}\right)\\- \frac{\partial}{\partial x} 3 x y z + \frac{d}{d z} \left(- 3 e^{x}\right)\\- \frac{d}{d y} \left(- 3 e^{x}\right) + \frac{d}{d x} \left(- 2 \cos{\left(z \right)}\right)\end{matrix}\right] = \left[\begin{matrix}3 x z - 2 \sin{\left(z \right)}\\- 3 y z\\0\end{matrix}\right] \]
curl F, component by component.✓ Proved
Answer \( \nabla\cdot\mathbf{F} = 3 x y - 3 e^{x},\quad \nabla\times\mathbf{F} = \langle 3 x z - 2 \sin{\left(z \right)}, - 3 y z, 0 \rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the divergence and curl of \( \displaystyle \mathbf{F} = \langle - 2 x y z, - 3 x, 2 \sin{\left(y \right)} \rangle \).
\[ \frac{d}{d y} \left(- 3 x\right) + \frac{\partial}{\partial x} \left(- 2 x y z\right) + \frac{d}{d z} 2 \sin{\left(y \right)} = - 2 y z \]
div F.✓ Proved
\[ \left[\begin{matrix}- \frac{d}{d z} \left(- 3 x\right) + \frac{d}{d y} 2 \sin{\left(y \right)}\\\frac{\partial}{\partial z} \left(- 2 x y z\right) - \frac{d}{d x} 2 \sin{\left(y \right)}\\\frac{d}{d x} \left(- 3 x\right) - \frac{\partial}{\partial y} \left(- 2 x y z\right)\end{matrix}\right] = \left[\begin{matrix}2 \cos{\left(y \right)}\\- 2 x y\\2 x z - 3\end{matrix}\right] \]
curl F, component by component.✓ Proved
Answer \( \nabla\cdot\mathbf{F} = - 2 y z,\quad \nabla\times\mathbf{F} = \langle 2 \cos{\left(y \right)}, - 2 x y, 2 x z - 3 \rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the divergence and curl of \( \displaystyle \mathbf{F} = \langle - x z, e^{x}, 3 x z \rangle \).
\[ \frac{\partial}{\partial x} \left(- x z\right) + \frac{\partial}{\partial z} 3 x z + \frac{d}{d y} e^{x} = 3 x - z \]
div F.✓ Proved
\[ \left[\begin{matrix}\frac{\partial}{\partial y} 3 x z - \frac{d}{d z} e^{x}\\\frac{\partial}{\partial z} \left(- x z\right) - \frac{\partial}{\partial x} 3 x z\\- \frac{\partial}{\partial y} \left(- x z\right) + \frac{d}{d x} e^{x}\end{matrix}\right] = \left[\begin{matrix}0\\- x - 3 z\\e^{x}\end{matrix}\right] \]
curl F, component by component.✓ Proved
Answer \( \nabla\cdot\mathbf{F} = 3 x - z,\quad \nabla\times\mathbf{F} = \langle 0, - x - 3 z, e^{x} \rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the divergence and curl of \( \displaystyle \mathbf{F} = \langle - x y z, - 3 z, 2 x y \rangle \).
\[ \frac{d}{d y} \left(- 3 z\right) + \frac{\partial}{\partial z} 2 x y + \frac{\partial}{\partial x} \left(- x y z\right) = - y z \]
div F.✓ Proved
\[ \left[\begin{matrix}- \frac{d}{d z} \left(- 3 z\right) + \frac{\partial}{\partial y} 2 x y\\- \frac{\partial}{\partial x} 2 x y + \frac{\partial}{\partial z} \left(- x y z\right)\\\frac{d}{d x} \left(- 3 z\right) - \frac{\partial}{\partial y} \left(- x y z\right)\end{matrix}\right] = \left[\begin{matrix}2 x + 3\\- x y - 2 y\\x z\end{matrix}\right] \]
curl F, component by component.✓ Proved
Answer \( \nabla\cdot\mathbf{F} = - y z,\quad \nabla\times\mathbf{F} = \langle 2 x + 3, - x y - 2 y, x z \rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the divergence and curl of \( \displaystyle \mathbf{F} = \langle 2 x^{2}, - 2 x, x y z \rangle \).
\[ \frac{d}{d y} \left(- 2 x\right) + \frac{d}{d x} 2 x^{2} + \frac{\partial}{\partial z} x y z = x y + 4 x \]
div F.✓ Proved
\[ \left[\begin{matrix}- \frac{d}{d z} \left(- 2 x\right) + \frac{\partial}{\partial y} x y z\\\frac{d}{d z} 2 x^{2} - \frac{\partial}{\partial x} x y z\\\frac{d}{d x} \left(- 2 x\right) - \frac{d}{d y} 2 x^{2}\end{matrix}\right] = \left[\begin{matrix}x z\\- y z\\-2\end{matrix}\right] \]
curl F, component by component.✓ Proved
Answer \( \nabla\cdot\mathbf{F} = x y + 4 x,\quad \nabla\times\mathbf{F} = \langle x z, - y z, -2 \rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the divergence and curl of \( \displaystyle \mathbf{F} = \langle - 2 x y, 2 x y, y \rangle \).
\[ \frac{d}{d z} y + \frac{\partial}{\partial x} \left(- 2 x y\right) + \frac{\partial}{\partial y} 2 x y = 2 x - 2 y \]
div F.✓ Proved
\[ \left[\begin{matrix}\frac{d}{d y} y - \frac{\partial}{\partial z} 2 x y\\- \frac{d}{d x} y + \frac{\partial}{\partial z} \left(- 2 x y\right)\\- \frac{\partial}{\partial y} \left(- 2 x y\right) + \frac{\partial}{\partial x} 2 x y\end{matrix}\right] = \left[\begin{matrix}1\\0\\2 x + 2 y\end{matrix}\right] \]
curl F, component by component.✓ Proved
Answer \( \nabla\cdot\mathbf{F} = 2 x - 2 y,\quad \nabla\times\mathbf{F} = \langle 1, 0, 2 x + 2 y \rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the divergence and curl of \( \displaystyle \mathbf{F} = \langle x y z, 2 x^{2}, y \rangle \).
\[ \frac{d}{d y} 2 x^{2} + \frac{d}{d z} y + \frac{\partial}{\partial x} x y z = y z \]
div F.✓ Proved
\[ \left[\begin{matrix}- \frac{d}{d z} 2 x^{2} + \frac{d}{d y} y\\- \frac{d}{d x} y + \frac{\partial}{\partial z} x y z\\\frac{d}{d x} 2 x^{2} - \frac{\partial}{\partial y} x y z\end{matrix}\right] = \left[\begin{matrix}1\\x y\\- x z + 4 x\end{matrix}\right] \]
curl F, component by component.✓ Proved
Answer \( \nabla\cdot\mathbf{F} = y z,\quad \nabla\times\mathbf{F} = \langle 1, x y, - x z + 4 x \rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Answer \( \nabla\cdot\mathbf{F} = 2 y,\quad \nabla\times\mathbf{F} = \langle 2 z, 0, - 2 x + 4 y \rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the divergence and curl of \( \displaystyle \mathbf{F} = \langle 3 y z, 2 x y z, z \rangle \).
\[ \frac{d}{d z} z + \frac{\partial}{\partial x} 3 y z + \frac{\partial}{\partial y} 2 x y z = 2 x z + 1 \]
div F.✓ Proved
\[ \left[\begin{matrix}\frac{d}{d y} z - \frac{\partial}{\partial z} 2 x y z\\- \frac{d}{d x} z + \frac{\partial}{\partial z} 3 y z\\- \frac{\partial}{\partial y} 3 y z + \frac{\partial}{\partial x} 2 x y z\end{matrix}\right] = \left[\begin{matrix}- 2 x y\\3 y\\2 y z - 3 z\end{matrix}\right] \]
curl F, component by component.✓ Proved
Answer \( \nabla\cdot\mathbf{F} = 2 x z + 1,\quad \nabla\times\mathbf{F} = \langle - 2 x y, 3 y, 2 y z - 3 z \rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the divergence and curl of \( \displaystyle \mathbf{F} = \langle - x y z, - 2 z^{2}, \sin{\left(y \right)} \rangle \).
\[ \frac{d}{d y} \left(- 2 z^{2}\right) + \frac{\partial}{\partial x} \left(- x y z\right) + \frac{d}{d z} \sin{\left(y \right)} = - y z \]
div F.✓ Proved
\[ \left[\begin{matrix}- \frac{d}{d z} \left(- 2 z^{2}\right) + \frac{d}{d y} \sin{\left(y \right)}\\\frac{\partial}{\partial z} \left(- x y z\right) - \frac{d}{d x} \sin{\left(y \right)}\\\frac{d}{d x} \left(- 2 z^{2}\right) - \frac{\partial}{\partial y} \left(- x y z\right)\end{matrix}\right] = \left[\begin{matrix}4 z + \cos{\left(y \right)}\\- x y\\x z\end{matrix}\right] \]
curl F, component by component.✓ Proved
Answer \( \nabla\cdot\mathbf{F} = - y z,\quad \nabla\times\mathbf{F} = \langle 4 z + \cos{\left(y \right)}, - x y, x z \rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the divergence and curl of \( \displaystyle \mathbf{F} = \langle 2 x y z, - 2 y^{2}, y \rangle \).
\[ \frac{d}{d z} y + \frac{d}{d y} \left(- 2 y^{2}\right) + \frac{\partial}{\partial x} 2 x y z = 2 y z - 4 y \]
div F.✓ Proved
\[ \left[\begin{matrix}\frac{d}{d y} y - \frac{d}{d z} \left(- 2 y^{2}\right)\\- \frac{d}{d x} y + \frac{\partial}{\partial z} 2 x y z\\\frac{d}{d x} \left(- 2 y^{2}\right) - \frac{\partial}{\partial y} 2 x y z\end{matrix}\right] = \left[\begin{matrix}1\\2 x y\\- 2 x z\end{matrix}\right] \]
curl F, component by component.✓ Proved
Answer \( \nabla\cdot\mathbf{F} = 2 y z - 4 y,\quad \nabla\times\mathbf{F} = \langle 1, 2 x y, - 2 x z \rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify