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Engineering Mathematics II • Digital Workbook • Page 7
Divergence and Solenoidal Fields
Divergence, Solenoidal Fields and the Laplacian
Divergence and solenoidal fields are developed through three IOE past-paper questions. Write each vector operator in full, differentiate carefully and follow the workbook method line by line.
Unit Vector Calculus
Topic Divergence
Question Type 2 Mark Problems
Theory Required for These Problems
Divergence of a Vector Field
For \(\mathbf F=P\mathbf i+Q\mathbf j+R\mathbf k\), divergence is the scalar
\[\mathrm{div}\,\mathbf F=\nabla\cdot\mathbf F=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}.\]
\[\mathrm{div}\,\mathbf F=\nabla\cdot\mathbf F\]\[\displaystyle =\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}.\]
Solenoidal Field
A vector field is solenoidal when its divergence is zero throughout the region:
\[\mathrm{div}\,\mathbf F=0.\]
Note: Differentiate only the matching component: \(P\) with respect to \(x\), \(Q\) with respect to \(y\), and \(R\) with respect to \(z\).
Gradient of a Scalar Function
Write the gradient in full before substituting derivatives:
\[\mathrm{grad}\,\phi=\nabla\phi=\mathbf i\frac{\partial\phi}{\partial x}+\mathbf j\frac{\partial\phi}{\partial y}+\mathbf k\frac{\partial\phi}{\partial z}.\]
\[\mathrm{grad}\,\phi=\nabla\phi\]\[\displaystyle =\mathbf i\frac{\partial\phi}{\partial x}+\mathbf j\frac{\partial\phi}{\partial y}+\mathbf k\frac{\partial\phi}{\partial z}.\]
Divergence of a Gradient
The divergence of the gradient is the Laplacian:
\[\mathrm{div}\,(\mathrm{grad}\,\phi)=\nabla^2\phi=\frac{\partial^2\phi}{\partial x^2}+\frac{\partial^2\phi}{\partial y^2}+\frac{\partial^2\phi}{\partial z^2}.\]
\[\mathrm{div}\,(\mathrm{grad}\,\phi)=\nabla^2\phi\]\[\displaystyle =\frac{\partial^2\phi}{\partial x^2}+\frac{\partial^2\phi}{\partial y^2}+\frac{\partial^2\phi}{\partial z^2}.\]
Method used: Identify the components, write the required operator in full, differentiate term by term, simplify, and then substitute the point when one is given.
23
Solenoidal Vector Field
2 Marks
PAST-PAPER REFERENCE: TU IOE • 2083 Baisakh • Back (New Course) • ENSH 151 • Q. 3(b) • 2 Marks
Prove that
\[\mathbf F=(y^2-z^2+3yz-2x)\mathbf i+(3xz+2xy)\mathbf j+(3xy-2xz+2z)\mathbf k\]
\[\mathbf F=(y^2-z^2+3yz-2x)\mathbf i\]\[+(3xz+2xy)\mathbf j\]\[+(3xy-2xz+2z)\mathbf k\]
is solenoidal.
Show Solution
Video Solution
Solution Divergence test
Write
\[\mathbf F=P\mathbf i+Q\mathbf j+R\mathbf k.\]
Here,
\[P=y^2-z^2+3yz-2x,\quad Q=3xz+2xy,\quad R=3xy-2xz+2z.\]
\[P=y^2-z^2+3yz-2x.\]\[Q=3xz+2xy.\]\[R=3xy-2xz+2z.\]
Now,
\[\mathrm{div}\,\mathbf F=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}.\]
Therefore,
\[\mathrm{div}\,\mathbf F=(-2)+(2x)+(-2x+2)=0.\]
\[\mathrm{div}\,\mathbf F=(-2)+(2x)+(-2x+2)\]\[=0.\]
Hence, the given vector field is solenoidal.
\[\mathrm{div}\,\mathbf F=0\]The vector field is solenoidal.
24
Divergence of a Gradient
2 Marks
PAST-PAPER REFERENCE: TU IOE • 2082 Bhadra • Regular (New Course) • ENSH 151 • Q. 3(b) • 2 Marks
If
\[\mathbf F=\mathrm{grad}\,(x^3+y^3+z^3-3xyz),\]
find \(\mathrm{div}\,\mathbf F\).
Show Solution
Video Solution
Solution Laplacian of a scalar function
Let
\[\phi=x^3+y^3+z^3-3xyz.\]
First write the gradient in full:
\[\mathrm{grad}\,\phi=\nabla\phi=\mathbf i\frac{\partial\phi}{\partial x}+\mathbf j\frac{\partial\phi}{\partial y}+\mathbf k\frac{\partial\phi}{\partial z}.\]
\[\mathrm{grad}\,\phi=\nabla\phi\]\[=\mathbf i\frac{\partial\phi}{\partial x}+\mathbf j\frac{\partial\phi}{\partial y}+\mathbf k\frac{\partial\phi}{\partial z}.\]
Since \(\mathbf F=\mathrm{grad}\,\phi\),
\[\mathrm{div}\,\mathbf F=\mathrm{div}\,(\mathrm{grad}\,\phi)=\nabla^2\phi.\]
Now,
\[\nabla^2\phi=\frac{\partial^2\phi}{\partial x^2}+\frac{\partial^2\phi}{\partial y^2}+\frac{\partial^2\phi}{\partial z^2}.\]
Differentiating twice,
\[\frac{\partial^2\phi}{\partial x^2}=6x,\quad\frac{\partial^2\phi}{\partial y^2}=6y,\quad\frac{\partial^2\phi}{\partial z^2}=6z.\]
\[\frac{\partial^2\phi}{\partial x^2}=6x.\]\[\frac{\partial^2\phi}{\partial y^2}=6y.\]\[\frac{\partial^2\phi}{\partial z^2}=6z.\]
Therefore,
\[\mathrm{div}\,\mathbf F=6x+6y+6z=6(x+y+z).\]
\[\mathrm{div}\,\mathbf F=6(x+y+z)\]
25
Divergence of Gradient at a Point
2 Marks
PAST-PAPER REFERENCE: TU IOE • 2081 Ashwin • Regular (New Course, 2080 Batch) • SH 151 • Q. 3(c) • 2 Marks
If
\[\phi=x^3+y^3+z^3-3xyz,\]
find \(\mathrm{div}\,(\mathrm{grad}\,\phi)\) at the point \((1,-1,1)\).
Show Solution
Video Solution
Solution Laplacian followed by substitution
Given
\[\phi=x^3+y^3+z^3-3xyz.\]
First write the gradient in full:
\[\mathrm{grad}\,\phi=\nabla\phi=\mathbf i\frac{\partial\phi}{\partial x}+\mathbf j\frac{\partial\phi}{\partial y}+\mathbf k\frac{\partial\phi}{\partial z}.\]
\[\mathrm{grad}\,\phi=\nabla\phi\]\[=\mathbf i\frac{\partial\phi}{\partial x}+\mathbf j\frac{\partial\phi}{\partial y}+\mathbf k\frac{\partial\phi}{\partial z}.\]
Now,
\[\mathrm{div}\,(\mathrm{grad}\,\phi)=\nabla^2\phi=\frac{\partial^2\phi}{\partial x^2}+\frac{\partial^2\phi}{\partial y^2}+\frac{\partial^2\phi}{\partial z^2}.\]
\[\mathrm{div}\,(\mathrm{grad}\,\phi)=\nabla^2\phi\]\[=\frac{\partial^2\phi}{\partial x^2}+\frac{\partial^2\phi}{\partial y^2}+\frac{\partial^2\phi}{\partial z^2}.\]
Therefore,
\[\nabla^2\phi=6x+6y+6z.\]
At \((1,-1,1)\),
\[\nabla^2\phi=6(1)+6(-1)+6(1)=6.\]
\[\mathrm{div}\,(\mathrm{grad}\,\phi)=6\quad \mathrm{at}\quad(1,-1,1)\]
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