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Dated: 30-10-2024

Scalar Fields

If a scalar field \(V(r)\) exists on all points of the curve then

\[\lim_{dr \to 0} \sum_{p = 1}^n V(r) dr_p = \int_c V(r) dr\]

Example

If \(V(x, y, z) = xy + y^2z\) then evaluate \(\int_c V(r)dr\) along the curve \(c\) defined by \(x = t^2\), \(y= 2t\) and \(z = t + 5\) between \(A(0, 0, 5)\) and \(B(4, 4, 7)\).

Solution

\[V(x, y, z) = xy + y^2z\]
\[\because x = t^2 \therefore dx = 2t \cdot dt\]
\[\because y = 2t \therefore dy = 2 \cdot dt\]
\[\because z = t + 5 \therefore dz = dt \]
\[V(t) = (t^2)(2t) + (2t)^2(t+5) = 60t^3 + 20t^2\]
\[d \vec r = dx \hat i + dy \hat j + dz \hat k\]
\[= 2t \cdot dt\cdot \hat i + 2 \cdot dt \cdot \hat j + dt \cdot \hat k\]
\[\therefore \int_c V(t) = \int_{C}(6t^{3}+20t^{2})(2t\hat {i}+2\hat{j}+\hat{k})dt\]

Integral1 Boundaries

Using the equations of \(x(t), y(t)\) and \(z(t)\).

\[A(0, 0, 5) \implies t = 0\]
\[B(4, 4, 7) \implies t = 2\]
\[\therefore \int_c V(t) = 2 \int_0^2 (6t^4 + 20t^3) \hat i + (6t^3 + 20t^2)\hat j + (3t^3 + 10t^2)\hat k \cdot dt\]
\[= \frac 8 {15} (444 \hat i + 290 \hat j + 145 \hat k)\]

Vector Field

Imagine we have a vector valued function2 \(\vec F(\vec r)\) for all points of a curve \(c\).
For each element of arc, we can have \(\vec F(t) \cdot d\vec r\) which gives us the length of that element.
Then the sum of all these lengths is

\[\sum_{p = 1}^n \vec F(\vec r) \cdot d\vec r_p\]

Therefore, the line integral3 for the whole curve is defined as

\[\lim_{d \vec r \to 0} \sum_{p = 1}^n \vec F(\vec r) \cdot d\vec r_p = \int_c \vec F(\vec r) \cdot d \vec r\]
\[= \int_c (F_1 \cdot dx + F_2 \cdot dy + F_3 \cdot dz)\]

Example

Evaluate \(\int_v \vec F dv\) where \(v\) is the region bounded by the planes4 \(x = 0\), \(y = 0\), \(z = 0\) and \(2x + y + z = 2\). Where \(\vec F = 2z \hat i + y \hat k\).

Solution

From the following, it is obvious that the region \(V\) lies in first octant.5

\[x = 0, y = 0, z = 0\]

Next, we will find where does the following plane4 intersects the space

\[2x + y + z = 2\]

Let us first find the points on the \(XY\) plane.4
For that, we will put \(z = 0\).

\[\therefore 2x + y = 2\]

We will now use this equation to find the \(x\) and \(y\) intercepts.6
For the \(x\) intercept, \(y = 0\)

\[\therefore x = 1\]

Similarly, for the \(y\) intercept, \(x = 0\)

\[\therefore y = 2\]

Now we will find the intercepts6 at the \(z\) axis.
For that, we will consider either \(XZ\) or \(YZ\) plane,4 any will do.
Putting \(y = 0\) for the \(XZ\) plane,4 we get

\[2x + z = 2\]

Put \(x = 0\) to find the \(z\) intercept6

\[\therefore z = 2\]

Therefore, the intercepts6 are \((1, 0, 0)\), \((0, 2, 0)\) and \((0, 0, 2)\).

\[\because 2x + y + z = 2\]
\[\therefore z = 2 - y - 2x\]
\[\int_v \vec F dv = \int_0^1 \int_0^{2(1 - x)}\int_0^{2(1 - x) - y} (2z\hat i + y \hat k) dzdydx\]
\[= \frac 1 3 (2 \hat i + \hat k)\]

References

Read more about notations and symbols.

  1. Read more about integrals. ↩

  2. Read more about vector valued functions. ↩

  3. Read more about line integral. ↩

  4. Read more about planes. ↩↩↩↩↩

  5. Read more about octants. ↩

  6. Read more about intercepts. ↩↩↩↩