Divergence and curl of a vector field at a given point. 44 problems with worked solutions; in 44 of them every
equation is proved by a computer algebra system.
20 worked problems
Try each one before opening the solution. Every line is labeled with what checked it. How we verify
Find the divergence and the curl of \( \displaystyle \mathbf F = \left\langle \sin{\left(y \right)}, \cos{\left(x \right)}, x y z \right\rangle \) at \( \displaystyle (1, 1, 1) \).
\[ \left. \frac{\partial}{\partial z} x y z + \frac{d}{d x} \sin{\left(y \right)} + \frac{d}{d y} \cos{\left(x \right)} \right|_{\substack{ x=1\\ y=1\\ z=1 }} = 1 \]
div F = P_x + Q_y + R_z, at the point.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial y} x y z - \frac{d}{d z} \cos{\left(x \right)} \right|_{\substack{ x=1\\ y=1\\ z=1 }}\\\left. - \frac{\partial}{\partial x} x y z + \frac{d}{d z} \sin{\left(y \right)} \right|_{\substack{ x=1\\ y=1\\ z=1 }}\\\left. - \frac{d}{d y} \sin{\left(y \right)} + \frac{d}{d x} \cos{\left(x \right)} \right|_{\substack{ x=1\\ y=1\\ z=1 }}\end{matrix}\right] = \left[\begin{matrix}1\\-1\\- \sin{\left(1 \right)} - \cos{\left(1 \right)}\end{matrix}\right] \]
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = 1,\quad \nabla\times\mathbf F = \left\langle 1, -1, - \sin{\left(1 \right)} - \cos{\left(1 \right)} \right\rangle \)
✓ 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 the curl of \( \displaystyle \mathbf F = \left\langle y e^{x}, x z, y^{2} \right\rangle \) at \( \displaystyle (1, 1, 1) \).
\[ \left. \frac{d}{d z} y^{2} + \frac{\partial}{\partial y} x z + \frac{\partial}{\partial x} y e^{x} \right|_{\substack{ x=1\\ y=1\\ z=1 }} = e \]
div F = P_x + Q_y + R_z, at the point.✓ Proved
\[ \left[\begin{matrix}\left. \frac{d}{d y} y^{2} - \frac{\partial}{\partial z} x z \right|_{\substack{ x=1\\ y=1\\ z=1 }}\\\left. - \frac{d}{d x} y^{2} + \frac{\partial}{\partial z} y e^{x} \right|_{\substack{ x=1\\ y=1\\ z=1 }}\\\left. \frac{\partial}{\partial x} x z - \frac{\partial}{\partial y} y e^{x} \right|_{\substack{ x=1\\ y=1\\ z=1 }}\end{matrix}\right] = \left[\begin{matrix}1\\0\\1 - e\end{matrix}\right] \]
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = e,\quad \nabla\times\mathbf F = \left\langle 1, 0, 1 - e \right\rangle \)
✓ 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
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = 3,\quad \nabla\times\mathbf F = \left\langle -1, -1, -1 \right\rangle \)
✓ 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 the curl of \( \displaystyle \mathbf F = \left\langle y z, x z, x y \right\rangle \) at \( \displaystyle (2, -1, 3) \).
\[ \left. \frac{\partial}{\partial z} x y + \frac{\partial}{\partial y} x z + \frac{\partial}{\partial x} y z \right|_{\substack{ x=2\\ y=-1\\ z=3 }} = 0 \]
div F = P_x + Q_y + R_z, at the point.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial y} x y - \frac{\partial}{\partial z} x z \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\\\left. - \frac{\partial}{\partial x} x y + \frac{\partial}{\partial z} y z \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\\\left. \frac{\partial}{\partial x} x z - \frac{\partial}{\partial y} y z \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\end{matrix}\right] = \left[\begin{matrix}0\\0\\0\end{matrix}\right] \]
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = 0,\quad \nabla\times\mathbf F = \left\langle 0, 0, 0 \right\rangle \)
✓ 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
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = 3,\quad \nabla\times\mathbf F = \left\langle -1, -1, -1 \right\rangle \)
✓ 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 the curl of \( \displaystyle \mathbf F = \left\langle x y, y z, x z \right\rangle \) at \( \displaystyle (2, -1, 3) \).
\[ \left. \frac{\partial}{\partial x} x y + \frac{\partial}{\partial z} x z + \frac{\partial}{\partial y} y z \right|_{\substack{ x=2\\ y=-1\\ z=3 }} = 4 \]
div F = P_x + Q_y + R_z, at the point.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial y} x z - \frac{\partial}{\partial z} y z \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\\\left. \frac{\partial}{\partial z} x y - \frac{\partial}{\partial x} x z \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\\\left. - \frac{\partial}{\partial y} x y + \frac{\partial}{\partial x} y z \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\end{matrix}\right] = \left[\begin{matrix}1\\-3\\-2\end{matrix}\right] \]
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = 4,\quad \nabla\times\mathbf F = \left\langle 1, -3, -2 \right\rangle \)
✓ 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 the curl of \( \displaystyle \mathbf F = \left\langle x y, y z, x z \right\rangle \) at \( \displaystyle (1, 2, -1) \).
\[ \left. \frac{\partial}{\partial x} x y + \frac{\partial}{\partial z} x z + \frac{\partial}{\partial y} y z \right|_{\substack{ x=1\\ y=2\\ z=-1 }} = 2 \]
div F = P_x + Q_y + R_z, at the point.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial y} x z - \frac{\partial}{\partial z} y z \right|_{\substack{ x=1\\ y=2\\ z=-1 }}\\\left. \frac{\partial}{\partial z} x y - \frac{\partial}{\partial x} x z \right|_{\substack{ x=1\\ y=2\\ z=-1 }}\\\left. - \frac{\partial}{\partial y} x y + \frac{\partial}{\partial x} y z \right|_{\substack{ x=1\\ y=2\\ z=-1 }}\end{matrix}\right] = \left[\begin{matrix}-2\\1\\-1\end{matrix}\right] \]
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = 2,\quad \nabla\times\mathbf F = \left\langle -2, 1, -1 \right\rangle \)
✓ 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 the curl of \( \displaystyle \mathbf F = \left\langle y e^{x}, x z, y^{2} \right\rangle \) at \( \displaystyle (2, -1, 3) \).
\[ \left. \frac{d}{d z} y^{2} + \frac{\partial}{\partial y} x z + \frac{\partial}{\partial x} y e^{x} \right|_{\substack{ x=2\\ y=-1\\ z=3 }} = - e^{2} \]
div F = P_x + Q_y + R_z, at the point.✓ Proved
\[ \left[\begin{matrix}\left. \frac{d}{d y} y^{2} - \frac{\partial}{\partial z} x z \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\\\left. - \frac{d}{d x} y^{2} + \frac{\partial}{\partial z} y e^{x} \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\\\left. \frac{\partial}{\partial x} x z - \frac{\partial}{\partial y} y e^{x} \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\end{matrix}\right] = \left[\begin{matrix}-4\\0\\3 - e^{2}\end{matrix}\right] \]
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = - e^{2},\quad \nabla\times\mathbf F = \left\langle -4, 0, 3 - e^{2} \right\rangle \)
✓ 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 the curl of \( \displaystyle \mathbf F = \left\langle x^{2} y, 2 y z, x z^{2} \right\rangle \) at \( \displaystyle (2, -1, 3) \).
\[ \left. \frac{\partial}{\partial z} x z^{2} + \frac{\partial}{\partial x} x^{2} y + \frac{\partial}{\partial y} 2 y z \right|_{\substack{ x=2\\ y=-1\\ z=3 }} = 14 \]
div F = P_x + Q_y + R_z, at the point.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial y} x z^{2} - \frac{\partial}{\partial z} 2 y z \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\\\left. - \frac{\partial}{\partial x} x z^{2} + \frac{\partial}{\partial z} x^{2} y \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\\\left. - \frac{\partial}{\partial y} x^{2} y + \frac{\partial}{\partial x} 2 y z \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\end{matrix}\right] = \left[\begin{matrix}2\\-9\\-4\end{matrix}\right] \]
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = 14,\quad \nabla\times\mathbf F = \left\langle 2, -9, -4 \right\rangle \)
✓ 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 the curl of \( \displaystyle \mathbf F = \left\langle x^{2} y, - y z, x z^{2} \right\rangle \) at \( \displaystyle (2, -1, 3) \).
\[ \left. \frac{\partial}{\partial z} x z^{2} + \frac{\partial}{\partial x} x^{2} y + \frac{\partial}{\partial y} - y z \right|_{\substack{ x=2\\ y=-1\\ z=3 }} = 5 \]
div F = P_x + Q_y + R_z, at the point.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial y} x z^{2} - \frac{\partial}{\partial z} - y z \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\\\left. - \frac{\partial}{\partial x} x z^{2} + \frac{\partial}{\partial z} x^{2} y \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\\\left. - \frac{\partial}{\partial y} x^{2} y + \frac{\partial}{\partial x} - y z \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\end{matrix}\right] = \left[\begin{matrix}-1\\-9\\-4\end{matrix}\right] \]
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = 5,\quad \nabla\times\mathbf F = \left\langle -1, -9, -4 \right\rangle \)
✓ 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
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = 3,\quad \nabla\times\mathbf F = \left\langle -1, -1, -1 \right\rangle \)
✓ 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 the curl of \( \displaystyle \mathbf F = \left\langle y e^{x}, x z, y^{2} \right\rangle \) at \( \displaystyle (1, 2, -1) \).
\[ \left. \frac{d}{d z} y^{2} + \frac{\partial}{\partial y} x z + \frac{\partial}{\partial x} y e^{x} \right|_{\substack{ x=1\\ y=2\\ z=-1 }} = 2 e \]
div F = P_x + Q_y + R_z, at the point.✓ Proved
\[ \left[\begin{matrix}\left. \frac{d}{d y} y^{2} - \frac{\partial}{\partial z} x z \right|_{\substack{ x=1\\ y=2\\ z=-1 }}\\\left. - \frac{d}{d x} y^{2} + \frac{\partial}{\partial z} y e^{x} \right|_{\substack{ x=1\\ y=2\\ z=-1 }}\\\left. \frac{\partial}{\partial x} x z - \frac{\partial}{\partial y} y e^{x} \right|_{\substack{ x=1\\ y=2\\ z=-1 }}\end{matrix}\right] = \left[\begin{matrix}3\\0\\- e - 1\end{matrix}\right] \]
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = 2 e,\quad \nabla\times\mathbf F = \left\langle 3, 0, - e - 1 \right\rangle \)
✓ 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 the curl of \( \displaystyle \mathbf F = \left\langle \sin{\left(y \right)}, \cos{\left(x \right)}, x y z \right\rangle \) at \( \displaystyle (1, 2, -1) \).
\[ \left. \frac{\partial}{\partial z} x y z + \frac{d}{d x} \sin{\left(y \right)} + \frac{d}{d y} \cos{\left(x \right)} \right|_{\substack{ x=1\\ y=2\\ z=-1 }} = 2 \]
div F = P_x + Q_y + R_z, at the point.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial y} x y z - \frac{d}{d z} \cos{\left(x \right)} \right|_{\substack{ x=1\\ y=2\\ z=-1 }}\\\left. - \frac{\partial}{\partial x} x y z + \frac{d}{d z} \sin{\left(y \right)} \right|_{\substack{ x=1\\ y=2\\ z=-1 }}\\\left. - \frac{d}{d y} \sin{\left(y \right)} + \frac{d}{d x} \cos{\left(x \right)} \right|_{\substack{ x=1\\ y=2\\ z=-1 }}\end{matrix}\right] = \left[\begin{matrix}-1\\2\\- \sin{\left(1 \right)} - \cos{\left(2 \right)}\end{matrix}\right] \]
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = 2,\quad \nabla\times\mathbf F = \left\langle -1, 2, - \sin{\left(1 \right)} - \cos{\left(2 \right)} \right\rangle \)
✓ 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 the curl of \( \displaystyle \mathbf F = \left\langle x y, y z, x z \right\rangle \) at \( \displaystyle (1, 1, 1) \).
\[ \left. \frac{\partial}{\partial x} x y + \frac{\partial}{\partial z} x z + \frac{\partial}{\partial y} y z \right|_{\substack{ x=1\\ y=1\\ z=1 }} = 3 \]
div F = P_x + Q_y + R_z, at the point.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial y} x z - \frac{\partial}{\partial z} y z \right|_{\substack{ x=1\\ y=1\\ z=1 }}\\\left. \frac{\partial}{\partial z} x y - \frac{\partial}{\partial x} x z \right|_{\substack{ x=1\\ y=1\\ z=1 }}\\\left. - \frac{\partial}{\partial y} x y + \frac{\partial}{\partial x} y z \right|_{\substack{ x=1\\ y=1\\ z=1 }}\end{matrix}\right] = \left[\begin{matrix}-1\\-1\\-1\end{matrix}\right] \]
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = 3,\quad \nabla\times\mathbf F = \left\langle -1, -1, -1 \right\rangle \)
✓ 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 the curl of \( \displaystyle \mathbf F = \left\langle x^{2} y, y z, x z^{2} \right\rangle \) at \( \displaystyle (0, 1, 2) \).
\[ \left. \frac{\partial}{\partial z} x z^{2} + \frac{\partial}{\partial x} x^{2} y + \frac{\partial}{\partial y} y z \right|_{\substack{ x=0\\ y=1\\ z=2 }} = 2 \]
div F = P_x + Q_y + R_z, at the point.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial y} x z^{2} - \frac{\partial}{\partial z} y z \right|_{\substack{ x=0\\ y=1\\ z=2 }}\\\left. - \frac{\partial}{\partial x} x z^{2} + \frac{\partial}{\partial z} x^{2} y \right|_{\substack{ x=0\\ y=1\\ z=2 }}\\\left. - \frac{\partial}{\partial y} x^{2} y + \frac{\partial}{\partial x} y z \right|_{\substack{ x=0\\ y=1\\ z=2 }}\end{matrix}\right] = \left[\begin{matrix}-1\\-4\\0\end{matrix}\right] \]
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = 2,\quad \nabla\times\mathbf F = \left\langle -1, -4, 0 \right\rangle \)
✓ 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 the curl of \( \displaystyle \mathbf F = \left\langle x^{2} y, y z, x z^{2} \right\rangle \) at \( \displaystyle (2, -1, 3) \).
\[ \left. \frac{\partial}{\partial z} x z^{2} + \frac{\partial}{\partial x} x^{2} y + \frac{\partial}{\partial y} y z \right|_{\substack{ x=2\\ y=-1\\ z=3 }} = 11 \]
div F = P_x + Q_y + R_z, at the point.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial y} x z^{2} - \frac{\partial}{\partial z} y z \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\\\left. - \frac{\partial}{\partial x} x z^{2} + \frac{\partial}{\partial z} x^{2} y \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\\\left. - \frac{\partial}{\partial y} x^{2} y + \frac{\partial}{\partial x} y z \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\end{matrix}\right] = \left[\begin{matrix}1\\-9\\-4\end{matrix}\right] \]
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = 11,\quad \nabla\times\mathbf F = \left\langle 1, -9, -4 \right\rangle \)
✓ 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 the curl of \( \displaystyle \mathbf F = \left\langle y z, x z, x y \right\rangle \) at \( \displaystyle (1, 1, 1) \).
\[ \left. \frac{\partial}{\partial z} x y + \frac{\partial}{\partial y} x z + \frac{\partial}{\partial x} y z \right|_{\substack{ x=1\\ y=1\\ z=1 }} = 0 \]
div F = P_x + Q_y + R_z, at the point.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial y} x y - \frac{\partial}{\partial z} x z \right|_{\substack{ x=1\\ y=1\\ z=1 }}\\\left. - \frac{\partial}{\partial x} x y + \frac{\partial}{\partial z} y z \right|_{\substack{ x=1\\ y=1\\ z=1 }}\\\left. \frac{\partial}{\partial x} x z - \frac{\partial}{\partial y} y z \right|_{\substack{ x=1\\ y=1\\ z=1 }}\end{matrix}\right] = \left[\begin{matrix}0\\0\\0\end{matrix}\right] \]
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = 0,\quad \nabla\times\mathbf F = \left\langle 0, 0, 0 \right\rangle \)
✓ 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 the curl of \( \displaystyle \mathbf F = \left\langle x^{2} y, - y z, x z^{2} \right\rangle \) at \( \displaystyle (1, 1, 1) \).
\[ \left. \frac{\partial}{\partial z} x z^{2} + \frac{\partial}{\partial x} x^{2} y + \frac{\partial}{\partial y} - y z \right|_{\substack{ x=1\\ y=1\\ z=1 }} = 3 \]
div F = P_x + Q_y + R_z, at the point.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial y} x z^{2} - \frac{\partial}{\partial z} - y z \right|_{\substack{ x=1\\ y=1\\ z=1 }}\\\left. - \frac{\partial}{\partial x} x z^{2} + \frac{\partial}{\partial z} x^{2} y \right|_{\substack{ x=1\\ y=1\\ z=1 }}\\\left. - \frac{\partial}{\partial y} x^{2} y + \frac{\partial}{\partial x} - y z \right|_{\substack{ x=1\\ y=1\\ z=1 }}\end{matrix}\right] = \left[\begin{matrix}1\\-1\\-1\end{matrix}\right] \]
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = 3,\quad \nabla\times\mathbf F = \left\langle 1, -1, -1 \right\rangle \)
✓ 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 the curl of \( \displaystyle \mathbf F = \left\langle x^{2} y, - 2 y z, x z^{2} \right\rangle \) at \( \displaystyle (2, -1, 3) \).
\[ \left. \frac{\partial}{\partial z} x z^{2} + \frac{\partial}{\partial x} x^{2} y + \frac{\partial}{\partial y} \left(- 2 y z\right) \right|_{\substack{ x=2\\ y=-1\\ z=3 }} = 2 \]
div F = P_x + Q_y + R_z, at the point.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial y} x z^{2} - \frac{\partial}{\partial z} \left(- 2 y z\right) \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\\\left. - \frac{\partial}{\partial x} x z^{2} + \frac{\partial}{\partial z} x^{2} y \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\\\left. - \frac{\partial}{\partial y} x^{2} y + \frac{\partial}{\partial x} \left(- 2 y z\right) \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\end{matrix}\right] = \left[\begin{matrix}-2\\-9\\-4\end{matrix}\right] \]
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = 2,\quad \nabla\times\mathbf F = \left\langle -2, -9, -4 \right\rangle \)
✓ 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 the curl of \( \displaystyle \mathbf F = \left\langle x^{2} y, y z, x z^{2} \right\rangle \) at \( \displaystyle (1, 2, -1) \).
\[ \left. \frac{\partial}{\partial z} x z^{2} + \frac{\partial}{\partial x} x^{2} y + \frac{\partial}{\partial y} y z \right|_{\substack{ x=1\\ y=2\\ z=-1 }} = 1 \]
div F = P_x + Q_y + R_z, at the point.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial y} x z^{2} - \frac{\partial}{\partial z} y z \right|_{\substack{ x=1\\ y=2\\ z=-1 }}\\\left. - \frac{\partial}{\partial x} x z^{2} + \frac{\partial}{\partial z} x^{2} y \right|_{\substack{ x=1\\ y=2\\ z=-1 }}\\\left. - \frac{\partial}{\partial y} x^{2} y + \frac{\partial}{\partial x} y z \right|_{\substack{ x=1\\ y=2\\ z=-1 }}\end{matrix}\right] = \left[\begin{matrix}-2\\-1\\-1\end{matrix}\right] \]
curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = 1,\quad \nabla\times\mathbf F = \left\langle -2, -1, -1 \right\rangle \)
✓ 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 ·
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