Dated: 30-10-2024
Examples
Example
Evaluate \(\int_v \vec F dv\) where \(\vec F = 2 \hat i + 2z \hat j + y \hat k\) and \(v\) is the region bounded by the planes \(z = 0, 4\) and the surface \(x^2 + y^2 = 9\).
Solution
In this chase, it would be convenient for us to convert the coordinates into polar coordinates.
\[\int_v \vec F dv = \iiint_v (2 \hat i + 2z \hat j + y \hat k) dxdydz\]
\[\because x = r \cdot \cos(\theta)\]
\[\because y = r \cdot \sin(\theta)\]
\[\because z = z\]
\[\because dv = r \cdot dr \cdot d\theta \cdot dz\]
\[= \int_0^{2 \pi} \int_0^3 \int_0^4 (2 \hat i + 2z \hat j + r \sin (\theta) \hat k) r \cdot dz \cdot dr \cdot \theta\]
\[= 72 \pi (\hat i + 2 \hat j)\]
Scalar Fields
A scalar field \(F = xyz\) exists over the surface \(S\) \(x^2 + y^2 = 4\) bounded by the planes \(z = 0, 3\).
Evaluate the following
\[\int_S F \cdot d \vec S\]
\[\because d \vec S = \hat n \cdot \lvert \vec S \rvert\]
\[\therefore \hat n = \frac {\nabla S}{\lvert \nabla S \rvert}\]
\[\nabla S = \hat i \frac{\partial S}{\partial x} + \hat j \frac{\partial S}{\partial y} + \hat k \frac{\partial S}{\partial z} = 2x \hat i + 2y \hat j\]
\[\lvert \nabla S \rvert = \sqrt{(2x)^2 + (2y)^2} = 2 \sqrt {x^2 + y^2} = 2 \sqrt 4 = 4\]
\[\therefore \hat n = \frac {x \hat i + y \hat j}{2}\]
\[\therefore d \vec S = \hat n \cdot dS\]
\[d \vec S = \frac {x \hat i + y \hat j}{2} dS\]
\[\int_S F \cdot d \vec S = \int_S F \cdot \hat n \cdot dS\]
\[= \int_S xyz \cdot \frac {x \hat i + y \hat j}{2} dS\]
\[=\frac{1}{2}\int_{S}(x^{2}yz \hat i+xy^{2}z \hat j) dS\]
Now we will convert the coordinates into polar coordinates.
\[\because x = r \cdot \cos(\theta) = 2 \cos (\theta)\]
\[\because y = r \cdot \sin(\theta) = 2 \sin(\theta)\]
\[\because z = z\]
\[\because dS = r \cdot dz \cdot d \theta = 2 \cdot dz \cdot d\theta\]
\[\because x^{2}yz=(4 \cos^{2}(\theta))(2 \sin(\theta))(z)=8 \cos^{2}(\theta)\sin(\theta) z\]
\[\because xy^{2}z=(2\cos(\theta))(4\sin^{2}(\theta))(z)=8\cos(\theta )\sin^{2}(\theta) z\]
\[\therefore \int_S F \cdot dS = \frac 8 2 \int_0^{\frac \pi 2} \int_0^3 (\cos^2(\theta) \sin(\theta)\hat i + \cos(\theta)\sin(2 \theta) \hat j) 2zdzd\theta\]
\[= 12 (\hat i + \hat j)\]
Vector Field
A vector field \(\vec F = y \hat i+ 2 \hat j + \hat k\) exists over the surface \(S\) defined by \(x^2 + y^2+ z^2 = 9\) and bounded by \(x = 0, y= 0, z= 0\) .
Evaluate
\[\int_S \vec F \cdot d \vec S\]
Solution
\[d \vec S = \hat n \cdot \lvert \vec S \rvert\]
\[\hat n = \frac{\nabla S}{\lvert \nabla S \rvert}\]
\[S: x^2 + y^2 + z^2 - 9 = 0\]
\[\nabla S = \frac{\partial S}{\partial x} \hat i + \frac{\partial S}{\partial y} \hat j + \frac{\partial S}{\partial z} \hat k = 2x \hat i + 2 \hat j + 2 \hat k\]
\[\lvert \nabla S\rvert=\sqrt{4x^{2}+4y^{2}+4z^{2}}=2\sqrt{x^{2}+y^{2}+z^{2}}=2\sqrt{9}=6\]
\[\therefore \hat n = \frac 1 3 (x \hat i + y \hat j + z \hat k)\]
\[\int_S \vec F \cdot d \vec S = \int_S \vec F \cdot \hat n \cdot dS\]
\[= \int_S (y \hat i + 2 \hat j + \hat k) \cdot \frac 1 3 (x \hat i + y \hat j + z \hat k) dS\]
\[ = \frac 1 3 \int_S (xy + 2y + z) dS\]
Before integrating over the surface, we will convert the coordinates into spherical coordinates.
\[x = 3 \sin (\phi) \cos (\theta)\]
\[y = 3 \sin (\phi) \sin (\theta)\]
\[z = 3 \cos(\phi)\]
\[dS = 9 \sin(\phi) d \phi d \theta\]
Both \(\phi\) and \(\theta\) are limited from \(0\) to \(\frac \pi 2\).
\[\because xy=3 \sin (\phi)\cos(\theta) \cdot 3 \sin(\phi) \sin(\theta) = 9 \sin^{2}(\theta)\cos(\theta)\]
\[\because 2y=2\cdot3 \sin(\phi)\sin(\theta)=6\sin(\phi)\sin(\theta)\]
\[\because z = 3 \cos (\phi)\]
\[\because dS = 9 \sin(\phi) d \phi d \theta\]
Putting these values in the integral, we get
\[\therefore \int_{S}\vec F \cdot d \vec S=\frac{1}{3}\int_{0}^{\frac \pi 2}\int_{0}^{\frac \pi 2}(9\sin^{2}(\phi) \sin(\theta) \cos(\theta)+6\sin(\phi) \sin(\theta)+3\cos(\phi))9\sin(\phi) d\phi d\theta\]
\[=9\left(1+\frac{3\pi}{4}\right)\]
References
Read more about notations and symbols.