Triple Integral and Coordinate Systems Study of Functions in Rectangular, Cylindrical, and Spherical Coordinates

Let $f$ be a continuous function on a solid region $Q$ .
The region $Q$ is defined by the following bounds: - $a \leq x \leq b$ - $h_{1}(x) \leq y \leq h_{2}(x)$ - $g_{1}(x, y) \leq z \leq g_{2}(x, y)$
In this definition, $h_{1}$ , $h_{2}$ , $g_{1}$ , and $g_{2}$ are continuous functions.
The triple integral of $f(x, y, z)$ over the region $Q$ is given by the iterated integral: - $\iiint_{Q} f(x, y, z) \,dV = \int_{a}^{b} \int_{h_{1}(x)}^{h_{2}(x)} \int_{g_{1}(x, y)}^{g_{2}(x, y)} f(x, y, z) \,dz \,dy \,dx$

1. Constant Multiple Rule: - $\iiint_{Q} c \,f(x, y, z) \,dV = c \iiint_{Q} f(x, y, z) \,dV$
2. Additive/Subtractive Rule: - $\iiint_{Q} [f(x, y, z) \pm g(x, y, z)] \,dV = \iiint_{Q} f(x, y, z) \,dV \pm \iiint_{Q} g(x, y, z) \,dV$
3. Union of Regions Rule: - If region $Q$ is the union of two non-overlapping regions $Q_{1}$ and $Q_{2}$ ( $Q = Q_{1} \cup Q_{2}$ ), then: - $\iiint_{Q} f(x, y, z) \,dV = \iiint_{Q_{1}} f(x, y, z) \,dV + \iiint_{Q_{2}} f(x, y, z) \,dV$

Problem: Evaluate the iterated integral $\int_{0}^{1} \int_{0}^{x} \int_{0}^{x+y} e^{x}(y + 2z) \,dz \,dy \,dx$ .
Step 1: Integrate with respect to $z$ : - $\int_{0}^{x+y} e^{x}(y + 2z) \,dz = [e^{x}(yz + z^{2})]_{0}^{x+y}$ - Substituting the limits: $e^{x}(y(x+y) + (x+y)^{2})$ - Expanding the terms: $e^{x}(xy + y^{2} + x^{2} + 2xy + y^{2})$ - Simplifying the expression: $e^{x}(x^{2} + 3xy + 2y^{2})$
Step 2: Integrate with respect to $y$ : - $\int_{0}^{x} e^{x}(x^{2} + 3xy + 2y^{2}) \,dy = e^{x}[x^{2}y + \frac{3}{2}xy^{2} + \frac{2}{3}y^{3}]_{0}^{x}$ - Substituting the limit $y = x$ : - $e^{x}(x^{2}(x) + \frac{3}{2}x(x)^{2} + \frac{2}{3}(x)^{3})$ - $e^{x}(x^{3} + \frac{3}{2}x^{3} + \frac{2}{3}x^{3})$ - Converting to a common denominator (6): $e^{x}(\frac{6}{6}x^{3} + \frac{9}{6}x^{3} + \frac{4}{6}x^{3}) = \frac{19}{6}x^{3}e^{x}$
Step 3: Integrate with respect to $x$ : - Finally, evaluate $\int_{0}^{1} \frac{19}{6}x^{3}e^{x} \,dx$ . - Using the anti-derivative for $x^{3}e^{x}$ : - $\frac{19}{6} [e^{x}(x^{3} - 3x^{2} + 6x - 6)]_{0}^{1}$ - Evaluating at the boundaries: - At $x = 1$ : $\frac{19}{6} e^{1}(1 - 3 + 6 - 6) = \frac{19}{6} e(-2)$ - At $x = 0$ : $\frac{19}{6} e^{0}(0 - 0 + 0 - 6) = \frac{19}{6} (-6)$ - Subtracting the results: $\frac{19}{6} (-2e - (-6)) = \frac{19}{6} (6 - 2e) = 19 - \frac{19}{3}e$ - The final numerical result provided is $\approx 65.797$ .

In the cylindrical coordinate system, a point $P$ in 3D space is represented by the ordered triple $(r, \theta, z)$ .
Definition of variables: - $r$ and $\theta$ are the polar coordinates of the projection of point $P$ onto the $xy$ -plane. - $z$ is the directed distance from the $xy$ -plane to point $P$ .
Conversion from Cylindrical to Rectangular coordinates: - $x = r \cos(\theta)$ - $y = r \sin(\theta)$ - $z = z$
Conversion from Rectangular to Cylindrical coordinates: - $r^{2} = x^{2} + y^{2}$ - $\tan(\theta) = \frac{y}{x}$ - $z = z$

Example 1: Cylindrical to Rectangular - Given cylindrical coordinates: $(2, 2\pi/3, 1)$ . - Find rectangular coordinates $(x, y, z)$ . - Solution: - $x = 2 \cos(2\pi/3) = 2(-\frac{1}{2}) = -1$ - $y = 2 \sin(2\pi/3) = 2(\frac{\sqrt{3}}{2}) = \sqrt{3}$ - $z = 1$ - The resulting point is $(-1, \sqrt{3}, 1)$ .
Example 2: Rectangular to Cylindrical - Given rectangular coordinates: $(3, -3, -7)$ . - Find cylindrical coordinates $(r, \theta, z)$ . - Solution: - $r = \sqrt{3^{2} + (-3)^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ - $\tan(\theta) = \frac{-3}{3} = -1$ - Note that the point is in Quadrant IV of the $xy$ -plane ( $x > 0$ , $y < 0$ ). - $\theta = \frac{7\pi}{4} + 2n\pi$ or $-\frac{\pi}{4}$ . - $z = -7$ - Therefore, one set of cylindrical coordinates is $(3\sqrt{2}, \frac{7\pi}{4}, -7)$ . Another set is $(3\sqrt{2}, -\frac{\pi}{4}, -7)$ .

The spherical coordinates of a point $P$ are captured as $(\rho, \theta, \phi)$ .
Geometric Interpretations: - $\rho = |OP|$ : The distance from the origin to point $P$ . Note $\rho \geq 0$ . - $\theta$ : The same angle as used in cylindrical and polar coordinates. - $\phi$ : The angle between the positive $z$ -axis and the line segment $OP$ . Bound by $0 \leq \phi \leq \pi$ .
Coordinate Relationships and Conversions: - $z = \rho \cos(\phi)$ - $r = \rho \sin(\phi)$ - Because $x = r \cos(\theta)$ and $y = r \sin(\theta)$ , the spherical to rectangular conversions are: - $x = \rho \sin(\phi) \cos(\theta)$ - $y = \rho \sin(\phi) \sin(\theta)$ - $z = \rho \cos(\phi)$ - The distance formula establishes the identity: $\rho^{2} = x^{2} + y^{2} + z^{2}$ .

Example 1: Spherical to Rectangular - Given spherical coordinates: $(2, \pi/4, \pi/3)$ . - Solution: - $x = \rho \sin(\phi) \cos(\theta) = 2 \sin(\pi/3) \cos(\pi/4) = 2(\frac{\sqrt{3}}{2})(\frac{\sqrt{2}}{2}) = \frac{\sqrt{6}}{2}$ - $y = \rho \sin(\phi) \sin(\theta) = 2 \sin(\pi/3) \sin(\pi/4) = 2(\frac{\sqrt{3}}{2})(\frac{\sqrt{2}}{2}) = \frac{\sqrt{6}}{2}$ - $z = \rho \cos(\phi) = 2 \cos(\pi/3) = 2(\frac{1}{2}) = 1$ - The resulting rectangular point is $(\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}, 1)$ .
Example 2: Rectangular to Spherical - Given rectangular coordinates: $(0, 2\sqrt{3}, -2)$ . - Solution: - $\rho = \sqrt{x^{2} + y^{2} + z^{2}} = \sqrt{0 + 12 + 4} = \sqrt{16} = 4$ - $\cos(\phi) = \frac{z}{\rho} = \frac{-2}{4} = -\frac{1}{2} \implies \phi = \frac{2\pi}{3}$ - Using $x = \rho \sin(\phi) \cos(\theta)$ : $0 = 4 \sin(\frac{2\pi}{3}) \cos(\theta)$ . Since the sine term is non-zero, $\cos(\theta) = 0$ , so $\theta = \frac{\pi}{2}$ . - The spherical coordinates are $(4, \pi/2, 2\pi/3)$ .

The formula for transforming a triple integral into spherical coordinates is: - $\iiint_{E} f(x, y, z) \,dV = \int_{c}^{d} \int_{\alpha}^{\beta} \int_{a}^{b} f(\rho \sin(\phi) \cos(\theta), \rho \sin(\phi) \sin(\theta), \rho \cos(\phi)) \rho^{2} \sin(\phi) \,d\rho \,d\theta \,d\phi$
This applies to a spherical wedge $E$ defined by: - $E = {(\rho, \theta, \phi) \mid a \leq \rho \leq b, \alpha \leq \theta \leq \beta, c \leq \phi \leq d}$

Problem: Evaluate $\iiint_{B} e^{(x^{2} + y^{2} + z^{2})^{3/2}} \,dV$ , where $B$ is the unit ball given by $B = {(x, y, z) \mid x^{2} + y^{2} + z^{2} \leq 1}$ .
Step 1: Identify Boundaries: - Since $B$ is a unit ball, the bounds in spherical coordinates are: - $0 \leq \rho \leq 1$ - $0 \leq \theta \leq 2\pi$ - $0 \leq \phi \leq \pi$
Step 2: Substitution: - Recognize that $x^{2} + y^{2} + z^{2} = \rho^{2}$ . The exponent becomes $(\rho^{2})^{3/2} = \rho^{3}$ .
Step 3: Setup the Integral: - $\int_{0}^{\pi} \int_{0}^{2\pi} \int_{0}^{1} e^{\rho^{3}} \rho^{2} \sin(\phi) \,d\rho \,d\theta \,d\phi$
Step 4: Evaluation: - Integrate with respect to $\rho$ : $\int_{0}^{1} \rho^{2} e^{\rho^{3}} \,drho = [\frac{1}{3} e^{\rho^{3}}]<em>{0}^{1} = \frac{1}{3}(e - 1)$ . - Integrate with respect to $\theta$ : $\int</em>{0}^{2\pi} d\theta = 2\pi$ . - Integrate with respect to $\phi$ : $\int_{0}^{\pi} \sin(\phi) \,d\phi = [-\cos(\phi)]_{0}^{\pi} = -(-1) - (-1) = 2$ .
Step 5: Final Result: - The product is $2\pi \times 2 \times \frac{1}{3}(e - 1) = \frac{4\pi}{3}(e - 1)$ .