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8.4: Double and Triple Integrals

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    106851
  • We can extend the idea of a definite integral to more dimensions. If \(f(x,y)\) is continuous over the rectangle \(R=[a,b]\times[c,d]\) then,

    \[\label{c2v:eq:doubleint} \underset{R}{\iint}f(x,y)dA=\int_{a}^{b}\int_{c}^{d}f{(x,y)\, dy\, dx}=\int_{c}^{d}\int_{a}^{b}f{(x,y)\, dx\, dy}\]

    If \(f(x,y)\geq0\), then the double integral represents the volume \(V\) of the solid that lies above the rectangle \(R\) and below the surface \(z=f(x,y)\) (Figure \(\PageIndex{1}\)).

    doubleintegral.jpg
    Figure \(\PageIndex{1}\): Geometric interpretation of a double integral (CC BY-NC-SA; Marcia Levitus)

    We can compute the double integral of Equation \ref{c2v:eq:doubleint} as:

    \[\underset{R}{\iint}f(x,y)dA=\int_{a}^{b}\left[ \int_{c}^{d}f{(x,y)\, dy} \right]dx \nonumber\]

    meaning that we will first compute

    \[\int_{c}^{d}f{(x,y)\, dy} \nonumber\]

    holding \(x\) constant and integrating with respect to \(y\). The result will be a function containing only \(x\), which we will integrate between \(a\) and \(b\) with respect to \(x\).

    For example, let’s solve \(\int_{0}^{3}\int_{1}^{2}{x^2 y\; dy\; dx}\). We’ll start by solving \(\int_{1}^{2}{x^2 y\; dy}\) holding \(x\) constant:

    \[\int_{1}^{2}{x^2 y\; dy}=x^2\int_{1}^{2}{y\; dy}=\dfrac{3}{2}x^2 \nonumber\]

    Now we integrate this function from 0 to 3 with respect to \(x\):

    \[\int_{0}^{3}\int_{1}^{2}{x^2 y\; dy\; dx}=\int_{0}^{3}{\dfrac{3}{2}x^2 \; dx}=\dfrac{27}{2} \nonumber\]

    You can of course integrate from 0 to 3 first with respect to \(x\) holding \(y\) constant, and then integrate the result with respect to \(y\) from 1 to 2. Try it this way and verify you get the same result.

    Triple integrals work in the same way. If \(f(x,y,z)\) is continuous on the rectangular box \(B=[a,b]\times[c,d]\times[r,s]\), then

    \[\label{c2v:eq:tripleint} \underset{B}{\iiint}f(x,y,z)dV=\int_{r}^{s}\int_{c}^{d}\int_{a}^{b}f{(x,y)\, dx\, dy\, dz}\]

    This iterated integral means that we integrate first with respect to \(x\) (keeping \(y\) and \(z\) fixed), then we integrate with respect to \(y\) (keeping \(z\) fixed), and finally we integrate with respect to \(z\). There are five other possible orders in which we can integrate, all of which give the same value.

    Do you need a refresher on double and triple integrals? Check the videos below before moving on to the physical chemistry examples.

    Triple integrals are used very often in physical chemistry to normalize probability density functions. For example, in quantum mechanics, the absolute square of the wave function, \(\left | \psi (x,y,z) \right |^2\), is interpreted as a probability density, the probability that the particle is inside the volume \(dx.dy.dz\). Since the probability of finding the particle somewhere in space is 1, we require that:

    \[\label{c2v:eq:calculus2v_normalization} \int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }{\left | \psi (x,y,z) \right |}^2\; dx \;dy \;dz=1\]

    We already mentioned wave functions in Section 2.3, where we showed that

    \[\left | \psi (x,y,z) \right |^2=\psi^*(x,y,z) \psi(x,y,z) \nonumber\]

    The normalization condition, therefore, can also be written as

    \[\label{c2v:eq:calculus2v_normalization2} \int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }{ \psi^*\psi }\; dx \;dy \;dz=1\]

    Example \(\PageIndex{1}\)

    In quantum mechanics, the lowest energy state of a particle confined in a three-dimensional box is represented by

    \[\psi (x,y,z)=A\sin{\dfrac{\pi x}{a}}\sin{\dfrac{\pi y}{b}}\sin{\dfrac{\pi z}{c}}\; if\; \left\{ \begin{matrix} 0< x< a\\ 0< y< b\\ 0< z< c\\ \end{matrix} \right. \]

    and

    \(\psi (x,y,z)=0\) otherwise (outside the box).

    Here, \(A\) is a normalization constant, and \(a\),\(b\) and \(c\) are the lengths of the sides of the box. Since the probability of finding the particle somewhere in space is 1, we require that

    \[\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }{\left | \psi (x,y,z) \right |}^2\; dx \;dy \;dz=1 \nonumber\]

    Find the normalization constant \(A\) in terms of \(a,b,c\) and other constants.

    Solution

    Because \(\psi(x,y,z)=0\) outside the box,

    \[\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }{\left | \psi (x,y,z) \right |}^2\; dx \;dy \;dz=\int_{0 }^{c }\int_{0 }^{b }\int_{0 }^{a }{\left | \psi (x,y,z) \right |}^2\; dx \;dy \;dz = \left | \psi (x,y,z) \right |^2=\psi^* (x,y,z) \psi(x,y,z) \nonumber\]

    However, because in this case the function is real,

    \[\left | \psi (x,y,z) \right |^2=\left ( \psi (x,y,z) \right )^2\]

    \[\int_{0 }^{c }\int_{0 }^{b }\int_{0 }^{a }{\left | \psi (x,y,z) \right |}^2\; dx \;dy \;dz=\int_{0 }^{c }\int_{0 }^{b }\int_{0 }^{a }{A^2 \sin^2\left( \dfrac{\pi x}{a}\right)\sin^2\left( \dfrac{\pi y}{b}\right)\sin^2\left( \dfrac{\pi z}{c}\right)}\; dx \;dy \;dz=1 \nonumber\]

    \[\int_{0 }^{c }\int_{0 }^{b }\int_{0 }^{a }{A^2 \sin^2\left( \dfrac{\pi x}{a}\right)\sin^2\left( \dfrac{\pi y}{b}\right)\sin^2\left( \dfrac{\pi z}{c}\right)}\; dx \;dy \;dz= A^2\int_{0 }^{a }{\sin^2\left( \dfrac{\pi x}{a}\right)}dx\int_{0 }^{b }{\sin^2\left( \dfrac{\pi y}{b}\right)}dy\int_{0 }^{c }{\sin^2\left( \dfrac{\pi z}{c}\right)}dz \nonumber\]

    Using the formula sheet, we get

    \[\int_{0 }^{a }{\sin^2\left( \dfrac{\pi x}{a}\right)}dx=a/2 \nonumber\]

    And therefore,

    \[A^2\int_{0 }^{a }{\sin^2\left( \dfrac{\pi x}{a}\right)}dx\int_{0 }^{b }{\sin^2\left( \dfrac{\pi y}{b}\right)}dy\int_{0 }^{c }{\sin^2\left( \dfrac{\pi z}{c}\right)}dz=A^2 \dfrac{a}{2} \dfrac{b}{2}\dfrac{c}{2}=\dfrac{A^2abc}{8}=1 \nonumber\]

    Solving for \(A\):

    \[\displaystyle{\color{Maroon}A=\left( \dfrac{8}{abc}\right)^{1/2}}\]

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