### Double and Triple Integrals

#### Double Integrals

##### Definition and Notation

The double integral of a function $z=f(x,y)$ represents the volume between the ($x,y$)-plane and the surface defined by that function.

This volume is given by $\iint\limits_R{f(x,y)} \; \mathrm{d} x\,\mathrm{d} y,$

where the projection of the function $z=f(x,y)$ on the ($x,y$)-plane defines the region $R$.

This is equivalent to the expression

$\int_{a}^{b}\int_{c(y)}^{d(y)} f(x,y) \; \mathrm{d} x\,\mathrm{d} y,$

where the region $R$ is defined by $\,c(y)A double integral can be treated as a repeated integral; the inner integral is computed first, followed by the outer one. In the above expression, the inner integral ###### Example 1 Evaluate$\displaystyle{ \int_1^2\int_{x}^{x^2} xy\;\mathrm{d} y\,\mathrm{d} x. }$###### Solution First evaluate the inner integral, treating$xas a constant: \begin{align} \int\limits_{x}^{x^2}xy\;\mathrm{d} y &= \left[x\dfrac{y^2}{2}\right]_x^{x^2} \\ &= x\cdot\dfrac{x^4}{2}-x\cdot\dfrac{x^2}{2} \\ &= \dfrac{x^5}{2}-\dfrac{x^3}{2} \end{align} Substitute this expression back into the original problem and evaluate the outer integral: \begin{align} \int_1^2{\left(\dfrac{x^5}{2}-\dfrac{x^3}{2}\right)} \; \mathrm{d} x &= \left[\dfrac{x^6}{12}-\dfrac{x^4}{8}\right]_1^2 \\ &= \left(\dfrac{64}{12}-\dfrac{16}{8}\right) - \left(\dfrac{1}{12}-\dfrac{1}{8}\right) \\ &= \dfrac{27}{8} \end{align} Hence we have shown that $\int_1^2\int_{x}^{x^2} xy\; \mathrm{d} y\,\mathrm{d} x=\frac{27}{8}.$ #### Triple Integrals Triple integrals are an extension of double integrals and are evaluated in the same way. Triple integrals often arise in calculations involving physical quantities, such as mass, volume or gravitational field. ##### Worked Example ###### Example 1 Use a triple integral to calculate the volume of a sphere of radiusR$. ###### Solution For calculations involving spherical objects it is convenient to use spherical polar coordinates$(r,\theta,\phi)$, where$r$is the radial distance,$\theta$is the azimuthal angle and$\phi$is the polar angle. For a sphere of radius$R$,$0\le r \le R$,$0 \le \theta \lt \pi$and$0 \le \phi \lt 2\pi$. Note that the volume element$\mathrm{d} V=\mathrm{d} x\,\mathrm{d} y\,\mathrm{d} z$becomes$\mathrm{d} V=r^2\sin{\theta}\,\mathrm{d} r\,\mathrm{d} \theta\,\mathrm{d} \phiwhen converted to spherical polars. Hence the triple integral required to calculate the volume of a sphere is $\text{Volume} = \iiint_V \mathrm{d} V = \int_0^{2\pi} \int_0^{\pi} \int_0^R r^2\sin{\theta}\,\mathrm{d} r\,\mathrm{d} \theta\,\mathrm{d} \phi$ To simplify the calculation, the order of the terms appearing within the integral can be rearranged: $\text{Volume} = \int_0^{2\pi} \int_0^{\pi} \int_0^R r^2 \mathrm{d} r\, \sin{\theta}\,\mathrm{d} \theta\,\mathrm{d} \phi$ First evaluate the inner integral: \begin{align} \int_0^R r^2 \mathrm{d} r &= \left[\dfrac{r^3}{3}\right]_0^R \\ &= \dfrac{R^3}{3} \end{align} Substitute this into the next innermost integral: $\int_0^{\pi} \dfrac{R^3}{3} \sin{\theta} \mathrm{d} \theta$ SinceRis a constant, it can be brought outside of the integral: \begin{align} \int_0^{\pi} \dfrac{R^3}{3} \sin{\theta}\, \mathrm{d} \theta &= \dfrac{R^3}{3} \int_0^{\pi} \sin{\theta} \mathrm{d} \theta \\ &= \dfrac{R^3}{3}\bigl[-\cos{\theta}\bigl]_0^{\pi} \\ &= \dfrac{R^3}{3}\left(1-(-1)\right) \\ &= 2\dfrac{R^3}{3} \end{align} Substitute this into the final, outer integral, again taking constants outside of the integral: \begin{align} \int_0^{2\pi} 2\dfrac{R^3}{3}\,\mathrm{d} \phi &= 2\dfrac{R^3}{3}\int_0^{2\pi} \mathrm{d} \phi \\ &= 2\dfrac{R^3}{3} \Bigl[\phi\Bigl]_0^{2\pi} \\ &= 2\dfrac{R^3}{3}(2\pi-0) \\ &=\dfrac{4}{3}\pi R^3 \end{align} Hence it has been shown that the volume of a sphere with radiusR\$ is

$\iiint_V \mathrm{d} V=\int_0^{2\pi} \int_0^{\pi} \int_0^R r^2\sin{\theta}\,\mathrm{d} r\,\mathrm{d} \theta\,\mathrm{d} \phi = \dfrac{4}{3}\pi R^3.$

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