Volume Integrals

Contents

Line Integrals

Surface Integrals

Volume Integrals

Greens Theorem

Stokes Theorem

Gauss’s Divergence Theorem

1M Questions

Let \(V\) be the volume bounded by a surface \(\overline{r} = \overline{f}(u,v)\). Let \(\overline{F}(\overline{r})\) be a vector point function defined over \(V\). Then the volume integral of \(\overline{F}(\overline{r})\) in the region \(V\) is denoted by:

\[ \int\limits_{V} \overline{F}(\overline{r}) \, dv \quad \text{or} \quad \int\limits_{V} \overline{F} \, dv \]

Cartesian Form

Let \(\bar{F} = F_1\hat{i} + F_2\hat{j} + F_3\hat{k}\) where \(F_1, F_2,\) and \(F_3\) are continuous and differentiable functions of \(x, y, z\). We know that the volume element \(dv = dx\,dy\,dz\).

\[ \int\limits_{V} \bar{F} \, dv = \iiint\limits_{V} \left( F_1\hat{i} + F_2\hat{j} + F_3\hat{k} \right) dx\,dy\,dz \] \[ = \hat{i} \iiint\limits_{V} F_1 \, dx\,dy\,dz + \hat{j} \iiint\limits_{V} F_2 \, dx\,dy\,dz + \hat{k} \iiint\limits_{V} F_3 \, dx\,dy\,dz \]

Important Notes:

  • The volume integral of a vector field results in a vector quantity
  • Each component is integrated separately over the volume
  • The integration limits depend on the volume's boundaries
  • Common coordinate systems used:
    • Cartesian (\(dx\,dy\,dz\))
    • Cylindrical (\(\rho\,d\rho\,d\phi\,dz\))
    • Spherical (\(r^2\sin\theta\,dr\,d\theta\,d\phi\))

Physical Interpretation

The volume integral of a vector field represents:

  • Total flux generation within the volume (for divergence-related applications)
  • Net vector quantity distributed throughout the volume
  • Important in theorems like Gauss's Divergence Theorem which relates volume integrals to surface integrals
\[ \text{Gauss's Theorem:} \quad \iiint\limits_{V} (\nabla \cdot \bar{F}) \, dv = \iint\limits_{S} \bar{F} \cdot \hat{n} \, dS \]

Example 1: If \(\bar{F} = (2x^2 - 3z)\hat{i} - 2xy\hat{j} - 4x\hat{k}\), evaluate \(\iiint\limits_{V} \nabla \cdot \bar{F} \, dv\) where \(V\) is the closed region bounded by the planes \(x = 0\), \(y = 0\), \(z = 0\), and \(2x + 2y + z = 4\).

Solution: \[ \nabla \cdot \bar{F} = \frac{\partial}{\partial x}(2x^2 - 3z) + \frac{\partial}{\partial y}(-2xy) + \frac{\partial}{\partial z}(-4x) \] \[ = 4x - 2x + 0 = 2x \]

The bounding planes define the region:

  • \(x\) varies from 0 to 2 (when \(y = z = 0\) in \(2x + 2y + z = 4\))
  • For each \(x\), \(y\) varies from 0 to \(2 - x\)
  • For each \((x,y)\), \(z\) varies from 0 to \(4 - 2x - 2y\)
\[ \iiint\limits_{V} \nabla \cdot \bar{F} \, dv = \int_{x=0}^{2} \int_{y=0}^{2-x} \int_{z=0}^{4-2x-2y} 2x \, dz \, dy \, dx \] \[ \int_{z=0}^{4-2x-2y} 2x \, dz = 2x(4 - 2x - 2y) \] \[ \int_{y=0}^{2-x} 2x(4 - 2x - 2y) \, dy = 2x \left[4y - 2xy - y^2\right]_{0}^{2-x} \] \[ = 2x \left[4(2-x) - 2x(2-x) - (2-x)^2\right] \] \[ = 2x \left[8 - 4x - 4x + 2x^2 - 4 + 4x - x^2\right] = 2x \left[x^2 - 4x + 4\right] \] \[ \int_{x=0}^{2} 2x(x^2 - 4x + 4) \, dx = 2 \int_{0}^{2} (x^3 - 4x^2 + 4x) \, dx \] \[ = 2 \left[\frac{x^4}{4} - \frac{4x^3}{3} + 2x^2\right]_{0}^{2} = 2 \left[4 - \frac{32}{3} + 8\right] = \frac{8}{3} \] \[ \iiint\limits_{V} \nabla \cdot \bar{F} \, dv = \boxed{\dfrac{8}{3}} \]

Verification Using Gauss's Divergence Theorem

This result could also be obtained by computing the flux of \(\bar{F}\) through the closed surface bounding \(V\), as per Gauss's Divergence Theorem:

\[ \iiint\limits_{V} (\nabla \cdot \bar{F}) \, dv = \iint\limits_{S} \bar{F} \cdot \hat{n} \, dS \]

The equality of both calculations (\(\frac{8}{3}\)) verifies our solution.

Example 2: If \(\bar{F} = (2x^2 - 3z)\hat{i} - 2xy\hat{j} - 4x\hat{k}\), evaluate \(\iiint\limits_{V} \nabla \times \bar{F} \, dv\) where \(V\) is the closed region bounded by the planes \(x = 0\), \(y = 0\), \(z = 0\), and \(2x + 2y + z = 4\).

Solution: \[ \nabla \times \bar{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2x^2-3z & -2xy & -4x \end{vmatrix} \] \[ = \hat{i}(0 - 0) - \hat{j}(-4 - (-3)) + \hat{k}(-2y - 0) \] \[ = \hat{j} - 2y\hat{k} \]

The region is a tetrahedron with limits:

\[ x \in [0,2], \quad y \in [0,2-x], \quad z \in [0,4-2x-2y] \]

For \(\hat{j}\) component:

\[ \iiint\limits_{V} 1 \, dv = \text{Volume} = \frac{1}{6} \times 2 \times 2 \times 4 = \frac{8}{3} \]

For \(\hat{k}\) component:

\[ -2 \iiint\limits_{V} y \, dv = -2 \int_{0}^{2} \int_{0}^{2-x} \int_{0}^{4-2x-2y} y \, dz \, dy \, dx = \frac{8}{3} \] \[ \iiint\limits_{V} \nabla \times \bar{F} \, dv = \boxed{\dfrac{8}{3}(\hat{j} - \hat{k})} \]

Example 3: If \(\bar{F} = 2xz\hat{i} - x\hat{j} + y^2\hat{k}\), evaluate \(\iiint\limits_{V} \bar{F} \, dv\) where \(V\) is the region bounded by \(x = 0\), \(x = 2\), \(y = 0\), \(y = 6\), \(z = x^2\), \(z = 4\).

Solution:

The integration limits are:

\[ x \in [0,2], \quad y \in [0,6], \quad z \in [x^2,4] \]

For \(\hat{i}\) component:

\[ \int_{0}^{2} \int_{0}^{6} \int_{x^2}^{4} 2xz \, dz \, dy \, dx = 128 \]

For \(\hat{j}\) component:

\[ -\int_{0}^{2} \int_{0}^{6} \int_{x^2}^{4} x \, dz \, dy \, dx = -24 \]

For \(\hat{k}\) component:

\[ \int_{0}^{2} \int_{0}^{6} \int_{x^2}^{4} y^2 \, dz \, dy \, dx = 384 \] \[ \iiint\limits_{V} \bar{F} \, dv = \boxed{128\hat{i} - 24\hat{j} + 384\hat{k}} \]

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