The divergence of the vector field \(3xz\hat i + 2xy\hat j - y{z^2}\hat k\) at a point (1,1,1) is equal to
3
Divergence Calculation Explained
Understanding the divergence of a vector field is a fundamental concept in vector calculus. The divergence measures the outward flux per unit volume from an infinitesimal volume around a point. It essentially tells us how much a vector field is "spreading out" or "converging" at a given point.
Vector Field Definition
The given vector field is denoted as \(\vec{F}\). It has three components, one for each spatial dimension (x, y, z):
\[ \vec{F} = P\hat{i} + Q\hat{j} + R\hat{k} \]
From the question, the components of the vector field are:
- P (component along \(\hat{i}\)): \(3xz\)
- Q (component along \(\hat{j}\)): \(2xy\)
- R (component along \(\hat{k}\)): \(-yz^2\)
Divergence Formula
The divergence of a vector field \(\vec{F} = P\hat{i} + Q\hat{j} + R\hat{k}\) is calculated using the following formula:
\[ \nabla \cdot \vec{F} = \frac{{\partial P}}{{\partial x}} + \frac{{\partial Q}}{{\partial y}} + \frac{{\partial R}}{{\partial z}} \]
This formula involves taking the partial derivative of each component with respect to its corresponding spatial variable and then summing these derivatives.
Partial Derivative Steps
Let's calculate each partial derivative step-by-step:
- Partial derivative of P with respect to x:
The component \(P = 3xz\). When taking the partial derivative with respect to \(x\), we treat \(z\) as a constant. \[ \frac{{\partial P}}{{\partial x}} = \frac{{\partial}}{{\partial x}}(3xz) = 3z \] - Partial derivative of Q with respect to y:
The component \(Q = 2xy\). When taking the partial derivative with respect to \(y\), we treat \(x\) as a constant. \[ \frac{{\partial Q}}{{\partial y}} = \frac{{\partial}}{{\partial y}}(2xy) = 2x \] - Partial derivative of R with respect to z:
The component \(R = -yz^2\). When taking the partial derivative with respect to \(z\), we treat \(y\) as a constant. \[ \frac{{\partial R}}{{\partial z}} = \frac{{\partial}}{{\partial z}}(-yz^2) = -2yz \]
Combining Derivatives for Divergence
Now, we sum these partial derivatives to find the general expression for the divergence of the vector field:
\[ \nabla \cdot \vec{F} = 3z + 2x - 2yz \]
Point Evaluation for Divergence
We need to find the divergence of the vector field at the specific point (1,1,1). To do this, we substitute \(x=1\), \(y=1\), and \(z=1\) into the divergence expression:
\[ \nabla \cdot \vec{F} \text{ at } (1,1,1) = 3(1) + 2(1) - 2(1)(1) \]
Let's perform the arithmetic:
- \(3 \times 1 = 3\)
- \(2 \times 1 = 2\)
- \(-2 \times 1 \times 1 = -2\)
So, the divergence at the point (1,1,1) is:
\[ 3 + 2 - 2 = 5 - 2 = 3 \]
Summary Table of Calculations
| Component | Partial Derivative | Value at (1,1,1) |
|---|---|---|
| \(P = 3xz\) | \(\frac{{\partial P}}{{\partial x}} = 3z\) | \(3(1) = 3\) |
| \(Q = 2xy\) | \(\frac{{\partial Q}}{{\partial y}} = 2x\) | \(2(1) = 2\) |
| \(R = -yz^2\) | \(\frac{{\partial R}}{{\partial z}} = -2yz\) | \(-2(1)(1) = -2\) |
Total Divergence = \(3 + 2 + (-2) = 3\)
Final Divergence Result
The divergence of the vector field \(3xz\hat i + 2xy\hat j - y{z^2}\hat k\) at the point (1,1,1) is 3.
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