3 Multiple Integrals
3.1 Multiple Integrals
For functions of multiple variables we may also consider integrals of multiple variables.
3.2 Integration (Review)
\(I=\int_a^b f(x) d x\) can be thought of as the area under the function \(f(x)\) from \(f(a)\) to \(f(b)\).
In this case \(I\) is a definite integral of \(f(x)\) between the lower limit \(x=a\) and upper limit \(x=b\), and \(f(x)\) is the integrand.
The definition of an integral as the area under the curve is not a formal definition - but it can be visualised, The formal definition of \(I\) involves subdividing the finite interval \(a \leq x \leq b\) into a large number of subintervals \(a=\xi_0<\xi_1<\xi_2<\ldots<\xi_n=b\) and then forming a sum \(\xi=\sum_{i=1}^n f\left(x_i\right)\left(\xi_i-\xi_{i-1}\right)\) where \(x_i\) is an arbitrary point that lies in the range \(\xi_{i-1} \leq x_i \leq \xi_i\). If \(n\) is now allowed to tend to infinity and the interval \(\xi_{i-1}\) to \(\xi_i\) tends to zero then \(S\) may or may not tend to a unique limit \(I\). If it does then the definite integral \(f(x)\) between \(a\) and \(b\) is defined as having the value \(I\). If no unique limit exists the integral is undefined. For continuous functions and finite intervals \(a \leq x \leq b\) the existence of a unique livest is assumed and the integral exists.
Evaluate from first principles the integral \[ I=\int_0^b x^2 d x \]
Lets first approximate the area under the curve \(x^2\) as a series of rectangles with width \(h\). If we the the lower end of each subinterval to give the height then the area of the \(k^{\text {th }}\) rectangle will be \[ \begin{aligned} & (k h)^2 \cdot h=k^2 h^3 \\ & A=\sum_{n=0}^{n-1} k^2 h^3=h^3 \frac{1}{6} n(n-1)(2 n-1) \quad \text { (sum of squares of natural numbers) } \\ & n=b / h \\ & A=\frac{b^3}{n^3} \cdot \frac{1}{6} n(n-1)(2 n-1) \\ & =\frac{b^3}{6 n^2}(n-1)(2 n-1) \\ & =\frac{b^3}{6}\left(1-\frac{1}{n}\right)(2-1 / n) \\ & \text { as } n \rightarrow \infty \quad A \rightarrow \frac{b^3}{3} \quad \therefore I=\frac{b^3}{3} . \end{aligned} \]
3.3 Double Integrals
For an integral involving two variables - a double integral \(f(x, y)\) to be integrated with respect to \(x\) and \(y\) between certain limits. These limits may represent a closed curve \(C\) bounding a region \(R\) in the \(xy\)-plane. In a similiar way to what we did with single integrals lets divide up the region \(R\) into \(N\) subregions \(\Delta R_p\) of are \(\Delta A_p, p=1,2, \ldots N\) and let \((x_p,y_p)\) be any point in subregion \(\Delta R_1\). Now consider the sum \[ S=\sum_{p=1}^N f\left(x_p, y_p\right) \Delta A_p \] and let \(N \rightarrow \infty\) as each of the areas \(\Delta A, \rightarrow 0\). If the sum \(S\) tends to a unique limit \(I\) then the is called the double integral of \(f(x, y)\) over the region \(R\) and is written \[ I=\int_R f(x, y) d A \]
Where \(d A\) stands for the element of area in the \(x y\)-plane. By choosing the subregions to be small rectangles each of are \(\Delta A=\Delta x \Delta y\) and letting both \(\Delta x\) and \(\Delta y \rightarrow 0\) we can also write the integral as \[ I=\iint_R f(x, y)\, d x d y \] where we here written out the area explicitly as the product of the coordinate differentials.
Note: Some authors use one integral no matter how many dimensions, some use one for each dimension. We will use the number of differentials explicitly written.
We can solve the integral above in more than one way - we can integrate over \(y\) first and then \(x\) or we can integrate over \(x\) first and then \(y\). If we do the first option we treat \(x\) as a constant first and integrate over the \(y\) for a specific \(x\). Here is how that is written: \[ I = \int_{x=a}^{x=b}\left(\int_{y=y_1(x)}^{y=y_2(x)}f(x,y)\,dy\right)\,dx \] The second option is the following: \[ I = \int_{y=ac}^{y=d}\left(\int_{x=x_1(y)}^{x=x_2(y)}f(x,y)\,dx\right)\,dy \]
These two options effectively change the order of the integration.
Evaluate the double integral \[ I = \int \int x^2 y \, dx dy, \] where \(R\) is the triangular region bounded by lines \(x=0\), \(y=0\) and \(x+y = 1\).
At each \(y\) integrate over \(x\) - this means horizontal chunks: \[ \begin{aligned} I &= \int_{y=0}^{y=1}\int_{x=0}^{x=1-y} x^2 y \, dx dy\\ &= \left.\int_0^1 \frac{x^3}{3}\right|_{x=0}^{x=1-y}y dy\\ &= \int_0^1 \frac{y(1-y)^3}{3} dy\\ &= \left. \frac{y^2}{6} - \frac{y^3}{3} + \frac{y^4}{4} - \frac{y^5}{15} \right|_0^1\\ &= \frac{1}{60} \end{aligned} \]
Now reverse the order of integration and show that the same result stands.
At each \(x\) integrate over \(y\) - this is vertical chunks. \[ \begin{aligned} I &= \int_{x=0}^{x=1}\int_{y=0}^{y=1-x} x^2 y \,dy dx\\ &= \left.\int_{x=0}^{x=1} \frac{x^2y^2}{2}\right|^{y=1-x}_{y=0} dx \\ &= \int_{x=0}^{x=1} \frac{x^2}{2}(1-x)^2 dx \\ &= \int_0^1 \frac{x^2}{2} - x^3 + \frac{x^4}{2} dx \\ &= \left.\left(\frac{x^3}{6} - \frac{x^4}{4} + \frac{x^5}{10}\right)\right|_0^1\\ &= \frac{1}{60} \end{aligned} \]
We can also write a double integral with the differential next to the integral: \[ I = \int_a^b dx \int_{y_1(x)}^{y_2(x)}dy\, f(x,y) \] where it is understood that each integral symbol acts on everything to its right and the order of integration is right to left. So \(f(x,y)\) is integrated with respect to \(y\) first and then \(x\).
When can the order of integrals not be changed?
If the region \(R\) is unbounded with some limits infinite or \(f(x,y)\) has any discontinuities in \(R\) or \(C\).
3.4 Triple Integrals
We can extend the same ideas from double integrals to triple integrals. Divide a region \(R\) into subregions \(\Delta R_p\) and volume \(\Delta V_p\) where \(p = 1, 2, ..., N\). The \(S\) is then: \[ = \sum_{p=1}^{N} f(x_p, y_p, z_p) \Delta V_p \] as \(N\rightarrow \infty\) \(\Delta V_p \rightarrow 0\) if \(S\) tends to a unique limit \(I\) then \[ I = \int_R f(x, y, z) \,dV \] where \(dV\) is an element of volume. \[ \begin{aligned} I &= \int\int\int_Rf(x, y,z)\, dx dy dz\\ &=\int_{x_1}^{x^2}dx\int_{y_1}^{y_2}dy\int_{z_1}^{z_2}dz\,f(x,y,z) \end{aligned} \]
3.5 Area and Volume
\[ A = \int_R dA = \int \int_R\,dx dy \] is equal to the area of region \(R\).
The volume under \(z= f(x,y)\) is \[ V = \int_R z \, dA = \int \int_R f(x,y)\, dx dy. \]
You can also find the volume directly by using a triple integral and integrating over \(z\).
Find the volume of the tetrahedron bounded by \(x=0\), \(y=0\), \(z=0\) and \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\).
It will help to visualize this if you draw a picture … \[ \begin{aligned} V &= \int \int_R z\, dx dy\\ z &= c(1-x/a - y/b)\\ V &= \int_0^a dx \int_0^{b(1-x/a)} dy c(1-x/a - y/b)\\ &= \left.\int_0^a dx c(y-xy/a - y^2/2b)\right|_0^{b-bx/a}\\ &= abc/6 \end{aligned} \]
3.6 Masses
Multiple integrals come in handy when we want to calculate the mass of an object with non-uniform density. \[ M = \int dM \] where \(dM\) is an element of mass and the integral is taken over the extent of the object. \[ dM = \rho \, dV \] where \(dV\) is an element of volume and \(\rho\) is the mass density.
Find the mass of the tetrahedron bounded by \(x=0\), \(y=0\), \(z=0\) and \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\) if the density is given by \(\rho(x,y,z) = \rho_o(1+x/a)\).
\[ \begin{aligned} M &= \int_R \rho_o(1+x/a)\,dV\\ &= \int_0^a dx \rho_o(1+x/a)\int_0^{b(1-x/a)} dy \int_0^{c(1-y/b-x/a)} dz\\ &= \frac{5}{24}\,abc\rho_o \end{aligned} \] where we have taken the density outside of the integrals with respect to \(y\) and \(z\) because it depends only on \(x\).
3.7 Centre of Mass
\[ \begin{aligned} \bar{x} &\int dM = \int x\, dM\\ \bar{y} &\int dM = \int y\, dM\\ \bar{z} &\int dM = \int z\, dM \end{aligned} \] where \(\int dM = M\).