Triple Integration
1. Introduction
Let $f(x,y,z)$ be a continuous function defined on a closed and bounded region $V \subset \mathbb{R}^3$. The triple integral of $f$ over $V$ is defined as
$$ \iiint_V f(x,y,z)\, dV $$
It represents the limit of Riemann sums:
$$ \iiint_V f(x,y,z)\, dV = \lim_{\max \Delta V_i \to 0} \sum f(x_i,y_i,z_i)\,\Delta V_i $$
provided the limit exists.
If $f(x,y,z)=1$, then the triple integral reduces to:
$$ \iiint_V 1\, dV = \text{Volume of } V $$
In practical computation, triple integrals are evaluated as iterated integrals:
$$ \iiint_V f(x,y,z)\, dV = \int_a^b \int_{g_1(x)}^{g_2(x)} \int_{h_1(x,y)}^{h_2(x,y)} f(x,y,z)\, dz\, dy\, dx $$
The order of integration may be changed whenever convenient.
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2. Explanation
Think of a double integral as adding up tiny rectangles to find the area of a region. A triple integral does the same thing in three dimensions — it adds up tiny boxes (small volumes) to find:
- Volume of a solid
- Mass (if density is given)
- Center of mass
- Physical quantities in engineering
Imagine breaking a solid object into many very small cubes. If we add up the value of the function at each cube times its small volume, we get the triple integral.
If we integrate just $1$, we simply get the volume:
$$ \iiint_V 1\, dV = \text{Volume} $$
3. Geometric Interpretation (3D Illustration)
Small Volume Elements Inside a Solid:
Projection of a Solid onto the xy-plane:
Common Coordinate Systems:
- Cartesian: $dV = dx\,dy\,dz$
- Cylindrical: $dV = r\,dr\,d\theta\,dz$
- Spherical: $dV = \rho^2 \sin\phi\, d\rho\, d\phi\, d\theta$
Summary
A triple integral allows us to measure quantities distributed throughout a three-dimensional region. It generalizes area (double integrals) to volume and physical applications in space.
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Triple Integrals – Solved Examples
Example 1
Evaluate:
$$ \int_{0}^{1}\int_{0}^{1-y}\int_{0}^{1-y-z} z \, dx \, dz \, dy $$Solution:
$$ = \int_{0}^{1}\int_{0}^{1-y} z(1-y-z)\, dz\, dy $$ $$ = \int_{0}^{1}\int_{0}^{1-y} \left[(1-y)z - z^2\right] dz\, dy $$ $$ = \int_{0}^{1} \left[ \frac{(1-y)z^2}{2} - \frac{z^3}{3} \right]_0^{1-y} dy $$ $$ = \int_{0}^{1} \left[ \frac{(1-y)^3}{2} - \frac{(1-y)^3}{3} \right] dy $$ $$ = \frac{1}{6}\int_{0}^{1}(1-y)^3 dy $$ $$ = \frac{1}{6}\left[\frac{(1-y)^4}{-4}\right]_0^1 $$ $$ = \frac{1}{24} $$Geometric Region:
Example 2
Evaluate:
$$ \iiint_V dV $$ where $$1 \le x \le 2,\quad 2 \le y \le 4,\quad 2 \le z \le 5.$$Solution:
$$ = \int_{2}^{4}\int_{1}^{2}\int_{2}^{5} dz\, dx\, dy $$ $$ = \int_{2}^{4}\int_{1}^{2} 3\, dx\, dy $$ $$ = \int_{2}^{4} 3(1)\, dy $$ $$ = 3[y]_2^4 $$ $$ = 6 $$Region (Rectangular Box):
Example 3
Evaluate:
$$ \iiint_V 2x\, dV $$ where $V$ lies under the plane $$2x + 3y + z = 6$$ in the first octant.Region Description:
$$ 0 \le z \le 6 - 2x - 3y $$ Projection on $xy$-plane: $$ 0 \le x \le 3, \quad 0 \le y \le \frac{6-2x}{3} $$Integral:
$$ \int_0^3 \int_0^{(6-2x)/3} 2x(6-2x-3y)\, dy\, dx $$ $$ = \int_0^3 \left(\frac{4}{3}x^3 - 8x^2 + 12x\right) dx $$ $$ = 9 $$Solid Figure:
Example 4
Evaluate:
$$ \iiint_V z(x^2+y^2)\, dV $$ over the cylinder $$x^2+y^2 \le 1$$ between $z=2$ and $z=3$.Integral Setup:
$$ = \int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \int_2^3 z(x^2+y^2)\, dz\, dx\, dy $$ $$ = \frac{5}{2}\int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} (x^2+y^2)\, dx\, dy $$ Using $y=\sin\theta$, $$ = \frac{10}{3}\int_0^{\pi/2} \cos^4\theta\, d\theta + 10\int_0^{\pi/2} \sin^2\theta\cos^2\theta\, d\theta $$ $$ = \frac{5\pi}{4} $$Cylinder Region:
Final Answers
- Example 2: 1/24
- Example 3: 6
- Example 4: 9
- Example 5: 5π/4
