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Engineering Mathematics II • Digital Workbook • Page 19
Gauss Divergence Theorem
Flux Across Closed Surfaces by Triple Integration
Gauss divergence theorem is developed through three IOE past-paper questions. Find the divergence, describe the closed volume and evaluate the required triple integral.
Unit Vector Calculus
Topic Gauss Theorem
Question Type 4 Mark Problems
Theory Required for These Problems
Gauss Divergence Theorem For a closed surface \(S\) enclosing a volume \(V\),
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS=\iiint_V\mathrm{div}\,\mathbf F\,dV.\]
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS\]\[=\iiint_V\mathrm{div}\,\mathbf F\,dV.\]
Divergence For \(\mathbf F=P\mathbf i+Q\mathbf j+R\mathbf k\), write
\[\mathrm{div}\,\mathbf F=\nabla\cdot\mathbf F=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}.\]
\[\mathrm{div}\,\mathbf F=\nabla\cdot\mathbf F\]\[=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}.\]
Choose the Volume Element For Cartesian bounds, use \(dV=dx\,dy\,dz\) in the required order. For a cylinder, use
\[dV=r\,dr\,d\theta\,dz.\]
Read the Closed Surface Identify the full enclosed volume before writing limits. A plane with the coordinate planes gives a first-octant tetrahedron.
Note: Gauss theorem applies to the complete closed surface and uses the outward normal.
Method used: Find \(\mathrm{div}\,\mathbf F=\nabla\cdot\mathbf F\), describe the enclosed volume, write the triple integral with correct bounds, and evaluate.
34
Flux Across a First-Octant Tetrahedron
4 Marks
PAST-PAPER REFERENCE: TU IOE • 2083 Baisakh • Back (New Course) • ENSH 151 • Q. 11 • 4 Marks
Apply the Gauss divergence theorem to evaluate
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS,\]
where
\[\mathbf F=(2xy+z^2)\mathbf i+y^2\mathbf j-(x+3y)\mathbf k,\]
\[\mathbf F=(2xy+z^2)\mathbf i+y^2\mathbf j\]\[-(x+3y)\mathbf k,\]
and \(S\) is the region bounded by
\[x+2y+z=6,\qquad x=0,\qquad y=0,\qquad z=0.\]
Show Solution
Video Solution
Solution Gauss theorem in Cartesian coordinates
The intercepts are (6) on the (x)-axis, (3) on the (y)-axis and (6) on the (z)-axis.
By Gauss divergence theorem,
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS=\iiint_V\mathrm{div}\,\mathbf F\,dV.\]
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS\]\[=\iiint_V\mathrm{div}\,\mathbf F\,dV.\]
Write the divergence in full:
\[\mathrm{div}\,\mathbf F=\nabla\cdot\mathbf F=\frac{\partial}{\partial x}(2xy+z^2)+\frac{\partial}{\partial y}(y^2)+\frac{\partial}{\partial z}[-(x+3y)].\]
\[\mathrm{div}\,\mathbf F=\nabla\cdot\mathbf F\]\[=\frac{\partial}{\partial x}(2xy+z^2)+\frac{\partial}{\partial y}(y^2)\]\[{}+\frac{\partial}{\partial z}[-(x+3y)].\]
Therefore,
\[\mathrm{div}\,\mathbf F=2y+2y+0=4y.\]
The bounds are
\[0\le x\le6,\qquad0\le y\le\frac{6-x}{2},\qquad0\le z\le6-x-2y.\]
\[0\le x\le6.\]\[0\le y\le\frac{6-x}{2}.\]\[0\le z\le6-x-2y.\]
Hence,
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS=\int_0^6\int_0^{(6-x)/2}\int_0^{6-x-2y}4y\,dz\,dy\,dx.\]
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS\]\[=\int_0^6\int_0^{(6-x)/2}\int_0^{6-x-2y}4y\,dz\,dy\,dx.\]
Integrating with respect to z,
\[=\int_0^6\int_0^{(6-x)/2}4y(6-x-2y)\,dy\,dx.\]
Integrating with respect to y,
\[=\int_0^6\frac{(6-x)^3}{6}\,dx.\]
Therefore,
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS=54.\]
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS=54\]
35
Flux Across a Parallelepiped
4 Marks
PAST-PAPER REFERENCE: TU IOE • 2082 Bhadra • Regular (New Course) • ENSH 151 • Q. 11 • 4 Marks
Use the Gauss divergence theorem to evaluate
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS,\]
where
\[\mathbf F=(x^2-yz)\mathbf i+(y^2-zx)\mathbf j+(z^2-xy)\mathbf k,\]
\[\mathbf F=(x^2-yz)\mathbf i\]\[+(y^2-zx)\mathbf j+(z^2-xy)\mathbf k,\]
and \(S\) is the surface of the parallelepiped bounded by
\[x=0,\quad x=a,\quad y=0,\quad y=b,\quad z=0,\quad z=c.\]
\[x=0,\qquad x=a,\qquad y=0,\qquad y=b.\]\[z=0,\qquad z=c.\]
Show Solution
Video Solution
Solution Gauss theorem over a rectangular volume
The closed surface encloses \(0\le x\le a\), \(0\le y\le b\) and \(0\le z\le c\).
By Gauss divergence theorem,
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS=\iiint_V\mathrm{div}\,\mathbf F\,dV.\]
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS\]\[=\iiint_V\mathrm{div}\,\mathbf F\,dV.\]
Write
\[\mathrm{div}\,\mathbf F=\nabla\cdot\mathbf F=\frac{\partial}{\partial x}(x^2-yz)+\frac{\partial}{\partial y}(y^2-zx)+\frac{\partial}{\partial z}(z^2-xy).\]
\[\mathrm{div}\,\mathbf F=\nabla\cdot\mathbf F\]\[=\frac{\partial}{\partial x}(x^2-yz)+\frac{\partial}{\partial y}(y^2-zx)\]\[{}+\frac{\partial}{\partial z}(z^2-xy).\]
Therefore,
\[\mathrm{div}\,\mathbf F=2x+2y+2z=2(x+y+z).\]
Hence,
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS=\int_0^a\int_0^b\int_0^c2(x+y+z)\,dz\,dy\,dx.\]
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS\]\[=\int_0^a\int_0^b\int_0^c2(x+y+z)\,dz\,dy\,dx.\]
Evaluating the three terms,
\[=a^2bc+ab^2c+abc^2.\]
Taking (abc) common,
\[=abc(a+b+c).\]
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS=abc(a+b+c).\]
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS\]\[=abc(a+b+c).\]
36
Flux Across a Closed Cylinder
4 Marks
PAST-PAPER REFERENCE: TU IOE • 2081 Ashwin • Regular (New Course, 2080 Batch) • SH 151 • Q. 11 • 4 Marks
Apply the Gauss divergence theorem to evaluate
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS\]
for
\[\mathbf F=x\mathbf i-y\mathbf j+(z^2-1)\mathbf k,\]
where \(S\) is the cylinder formed by
\[z=0,\qquad z=1,\qquad x^2+y^2=4.\]
Show Solution
Video Solution
Solution Gauss theorem in cylindrical coordinates
Use cylindrical coordinates with \(0\le r\le2\), \(0\le\theta\le2\pi\) and \(0\le z\le1\).
Write the divergence in full:
\[\mathrm{div}\,\mathbf F=\nabla\cdot\mathbf F=\frac{\partial x}{\partial x}+\frac{\partial(-y)}{\partial y}+\frac{\partial(z^2-1)}{\partial z}.\]
\[\mathrm{div}\,\mathbf F=\nabla\cdot\mathbf F\]\[=\frac{\partial x}{\partial x}+\frac{\partial(-y)}{\partial y}+\frac{\partial(z^2-1)}{\partial z}.\]
Therefore,
\[\mathrm{div}\,\mathbf F=1-1+2z=2z.\]
By Gauss divergence theorem,
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS=\iiint_V2z\,dV.\]
For the cylinder,
\[0\le r\le2,\qquad0\le\theta\le2\pi,\qquad0\le z\le1,\qquad dV=r\,dr\,d\theta\,dz.\]
\[0\le r\le2,\qquad0\le\theta\le2\pi.\]\[0\le z\le1,\qquad dV=r\,dr\,d\theta\,dz.\]
Hence,
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS=\int_0^1\int_0^{2\pi}\int_0^2 2zr\,dr\,d\theta\,dz.\]
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS\]\[=\int_0^1\int_0^{2\pi}\int_0^2 2zr\,dr\,d\theta\,dz.\]
Separating the factors,
\[=\left[ z^2\right]_0^1\left[\theta\right]_0^{2\pi}\left[\frac{r^2}{2}\right]_0^2.\]
\[=\left[ z^2\right]_0^1\left[\theta\right]_0^{2\pi}\]\[\left[\frac{r^2}{2}\right]_0^2.\]
Therefore,
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS=(1)(2\pi)(2)=4\pi.\]
\[\iint_S\mathbf F\cdot\widehat{\mathbf n}\,dS=4\pi\]
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