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Multivariable Calculus
by Samuel S. Watson
Introduction
Taylor Series
Differentiation
Optimization
Matrix differentiation
The chain rule
Multivariable integration
Custom integration

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Multivariable CalculusMultivariable integration

Lugemise aeg: ~15 min

Integrating a function is a way of totaling up its values. For example, if f is a function from a region D in \mathbb{R}^n to \mathbb{R} which represents the mass density of a solid occupying the region D, we can find the total mass of the solid as follows: (i) split the region D into many tiny pieces, (ii) multiply the volume of each piece by the value of the function at some point on that piece, (iii) and add up the results. If we take the number of pieces to infinity and the piece size to zero, then this sum converges to the total mass of the solid.

We may apply this procedure to any function f defined on D, and we call the result the integral of f over D, denoted \int_D f.

To find the integral of a function f defined on a 2D region D, we set up a double iterated integral over D: the bounds for the outer integral are the smallest and largest possible values of y for point in D, and the bounds for the inner integral are the smallest and largest values of x for any point in a given each "y = constant" slice of the region (assuming that each slice intersects the region in a line segment).

Exercise
Find the integral over the triangle T with vertices (0,0), (2,0), and (0,3) of the function f(x,y) = x^2y.

Solution. The least and greatest values of y for any point in the region are 0 and 3, while the least and greatest values of x for each given y-slice are 0 and 2 - \frac{2}{3}y. Therefore, the integral is

\begin{align*}\int_0^3 \int_0^{2 - \frac{2}{3}y} x^2 y \, \mathrm{d} x \, \mathrm{d} y = \frac{6}{5}.\end{align*}

3D integration

To set up an integral of a function over a 3D region (for the order \mathrm{d} x \, \mathrm{d}y \, \mathrm{d}z): the bounds for the outer integral are the smallest and largest values of z for any point in the region of integration, then the bounds for the middle integral are the smallest and largest values of y for any point in the region in each "z = constant" plane, and the inner bounds are the smallest and largest values of x for any point in the region in each "(y,z) = constant" line.

Exercise
Integrate the function f(x,y,z) = 1 over the tetrahedron with vertices (0,0,0), (2,0,0), (0,3,0), and (0,0,4).

Solution. The least and greatest values of z are 0 and 4, so those are our outer limits (see the figure below). For a fixed value of z, the least and greatest values of y for a point in D are 0 and 3 - \tfrac{3}{4}z, respectively. Finally, for fixed y and z, the least and greatest values of x for a point in D are 0 and the point on the plane 6x + 4y + 3z = 12 with the given values of y and z. So we get

\begin{align*}\text{volume}(D) &= \int_{0}^{4}\int_{0}^{3-\frac{3}{4}z}\int_{0}^{2 - \frac{2}{3}y - \frac{1}{2}z} 1 \, \mathrm{d}x \, \mathrm{d}y \, \mathrm{d}z \\\ &= \int_{0}^{4}\int_{0}^{3-\frac{3}{4}z} \left(2 - \frac{2}{3}y - \frac{1}{2}z \right) \, \mathrm{d}y \, \mathrm{d}z \\\ &= \int_{0}^{4} \frac{3}{16}(z-4)^2 \, \mathrm{d}z \\\ &= \boxed{4}.\end{align*}

Exercises

Exercise
Evaluate \int_0^8 \int_{\sqrt[3]{y}}^2 e^{x^4} , \mathrm{d} x ,\mathrm{d} y by writing it as an integral over a region in the plane and then integrating over the region with respect to the opposite order of integration.

Solution Let's begin by drawing the region:

We can see that this is the region under the graph of y = x^3 from x = 0 to x = 2. Thus we integrate as x ranges from 0 to 2 and (for each fixed value of x) as y ranges from 0 to x^3. We get

\begin{align*}\int^2_0 \int^{x^3}_0 e^{x^4}\,\mathrm{d} y\,\mathrm{d} x &= \int^2_0 \int^{x^3}_0 y e^{x^4} \bigg\rvert_{y=0}^{y={x^3}}\,\mathrm{d} x \\\ &= \int^2_0 x^3e^{x^4} \,\mathrm{d} x = \frac{e^{x^4}}{4}\bigg\rvert_0^2 = \frac{e^{16} - 1}{4} \approx 2221527.\end{align*}

Exercise
Consider the region R between the parabolas y=1-x^2 and y=x^2-7. Find \iint_R xy \, \mathrm{d} A.

Solution. We see that the curves intersect at x = -2 and x = 2. So we get

\begin{align*}\int^2_{-2} \int^{1-x^2}_{x^2-7} xy \,\mathrm{d} y\,\mathrm{d} x &= \int^{2}_{-2} {\frac{xy^2}{2} \bigg\rvert_{x^2-7}^{1-x^2}} \,\mathrm{d} x \\\ &= \int^2_{-2} {\frac{x}{2}[(1 - 2x^2 +x^4)-(x^4 -14x^2 +49)]} \,\mathrm{d} x \\\ &= \int^2_{-2} {6x^3 - 24x} \,\mathrm{d} x \\\ &= \frac{3x^4}{2}-12x^2 \bigg\rvert^2_{-2} \\\ &= 0\end{align*}

Bruno
Bruno Bruno