Partial Derivatives and Multiple Integrals

The MSTE paper moves from one variable to several the moment a formula has more than one input, and the tools change only a little. This lesson covers the four multivariable ideas the CELE actually tests: partial derivatives, the total differential and how it propagates measurement error, the gradient and the directional derivative, and the double integral used for area, volume, and centroids. Every rule is carried all the way to a number so you can check your own habits against it.

Partial derivatives

A partial derivative treats a function of several variables as if only one of them can move. To find partial f / partial x you differentiate f with respect to x while holding y (and any other variable) constant; for partial f / partial y you hold x constant. Nothing new is memorized: the power, product, quotient, and chain rules all still apply, and the other variables ride along as constants.

Symbol	Also written	Meaning
partial f / partial x	f_x	rate of change of f in the x-direction, y held constant
partial f / partial y	f_y	rate of change of f in the y-direction, x held constant
partial^2 f / partial x^2	f_xx	differentiate with respect to x twice
partial^2 f / partial y partial x	f_xy	differentiate with respect to x, then with respect to y
gradient grad f	(f_x, f_y)	the vector of the two first partials

Worked example (first partials): Let f(x, y) = x^2 y + 3 x y^2. Holding y constant, f_x = 2xy + 3y^2; holding x constant, f_y = x^2 + 6xy. At the point (1, 2), f_x = 2(1)(2) + 3(2^2) = 4 + 12 = 16 and f_y = 1^2 + 6(1)(2) = 1 + 12 = 13. Worked example (second partials): For the same f, differentiate f_x = 2xy + 3y^2 again in x to get f_xx = 2y, and in y to get the mixed partial f_xy = 2x + 6y. Starting instead from f_y = x^2 + 6xy gives f_yx = 2x + 6y, the same result. This equality, f_xy = f_yx whenever the second partials are continuous, is Clairaut's theorem, and it is a fast self-check: if your two mixed partials disagree, one of them is wrong.

The total differential and error propagation

Small changes in the inputs produce a small change in the output, and the total differential is the linear estimate of that change:

df = f_x dx + f_y dy.

Read it as a bookkeeping rule: each input contributes its own partial derivative times how much that input moved. This is exactly how a measured quantity carries its measurement error forward. If z = f(x, y) and x and y are each known only to within dx and dy, then dz = f_x dx + f_y dy estimates the resulting uncertainty in z. Dividing through by the quantity itself turns absolute error into relative (percent) error, which for a product of powers separates into a clean sum.

Worked example: The volume of a right circular cylinder is V = pi r^2 h, measured at r = 5 cm and h = 10 cm, so V = pi (5^2)(10) = 250 pi = 785.4 cm^3. Suppose the radius is uncertain by dr = 0.1 cm and the height by dh = 0.2 cm. The partials are partial V / partial r = 2 pi r h and partial V / partial h = pi r^2, so

dV = 2 pi r h dr + pi r^2 dh = 2 pi (5)(10)(0.1) + pi (5^2)(0.2) = 10 pi + 5 pi = 15 pi = 47.1 cm^3.

The relative error is cleaner still. Dividing dV by V = pi r^2 h gives dV / V = 2 dr / r + dh / h = 2(0.1/5) + (0.2/10) = 2(0.02) + 0.02 = 0.06, so the volume is uncertain by about 6 percent. Notice the radius enters with a factor of 2 because it appears squared in the formula, the single most useful pattern in error propagation.

The gradient and the directional derivative

Collect the first partials into one vector and you have the gradient, grad f = (f_x, f_y). It carries two pieces of information at once: it points in the direction of steepest ascent of f, and its length is the maximum rate of increase. It is also always perpendicular to the level curves f = c, so if you can sketch the level curves you already know which way the gradient points.

The directional derivative measures the rate of change of f in one chosen direction. For a unit vector u, it is the dot product

D_u f = grad f dot u.

Because a dot product is largest when the two vectors are aligned, the biggest possible directional derivative equals the length of the gradient, achieved when u points along grad f. Along a level curve the direction is perpendicular to the gradient, so the dot product is zero and f is momentarily flat.

Worked example: Let f(x, y) = x^2 + y^2, whose gradient is grad f = (2x, 2y). At the point (3, 4), grad f = (6, 8). The maximum rate of increase there is its magnitude, |grad f| = sqrt(6^2 + 8^2) = sqrt(36 + 64) = sqrt(100) = 10, and the steepest-ascent direction is the unit vector (6, 8)/10 = (0.6, 0.8). To find the rate of change toward the vector (1, 1), first normalize it: u = (1, 1)/sqrt(2). Then D_u f = (6, 8) dot (1/sqrt(2), 1/sqrt(2)) = (6 + 8)/sqrt(2) = 14/sqrt(2) = 7 sqrt(2) = 9.90 units per unit length. Forgetting to normalize u and reporting 14 is the classic slip.

Setting up a double integral

A double integral adds a quantity up over a two-dimensional region R. By Fubini's theorem you evaluate it as two ordinary integrals done one at a time, working from the inside out. For a Type I region (bounded left and right by fixed x-values and top and bottom by curves), you integrate dy first, holding x fixed, then dx. Geometrically the inner integral sweeps a single vertical strip from the lower boundary to the upper boundary, and the outer integral slides that strip across the region.

Worked example (area): Find the area between y = x^2 and y = 2x. The curves meet where 2x = x^2, that is x^2 - 2x = 0, so x = 0 or x = 2, and on that interval 2x sits above x^2. Written as a double integral the area is

A = integral from x = 0 to 2 of integral from y = x^2 to 2x of dy dx = integral from 0 to 2 of (2x - x^2) dx = [x^2 - x^3/3] from 0 to 2 = (4 - 8/3) = 4/3 square units.

Volume under a surface

When the integrand is a height function f(x, y) >= 0 rather than just dA, the double integral gives the volume of the solid trapped between the region R in the xy-plane and the surface z = f(x, y) above it. Each thin column has base area dA and height f(x, y), so its volume is f(x, y) dA, and the double integral totals them.

Worked example: Find the volume under the plane z = 8 - 2x - y over the rectangle 0 <= x <= 2, 0 <= y <= 3. Over this rectangle z stays positive (its smallest value, at x = 2 and y = 3, is 8 - 4 - 3 = 1). Integrate dy first:

integral from y = 0 to 3 of (8 - 2x - y) dy = [8y - 2xy - y^2/2] from 0 to 3 = 24 - 6x - 4.5 = 19.5 - 6x.

Then integrate that in x: integral from 0 to 2 of (19.5 - 6x) dx = [19.5x - 3x^2] from 0 to 2 = 39 - 12 = 27 cubic units. As a second drill, the volume under z = xy over the same rectangle is integral from 0 to 2 of integral from 0 to 3 of xy dy dx = integral from 0 to 2 of x(9/2) dx = (9/2)(2) = 9 cubic units.

Centroids by double integration

The double integral also locates the centroid of a plane region, the balance point of a uniform lamina. The first moment about an axis is the integral of the perpendicular distance to that axis, and dividing a moment by the area gives a coordinate of the centroid.

Quantity for region R	Double integral
Area A	double integral over R of dA
Volume under z = f(x, y) >= 0	double integral over R of f(x, y) dA
Mass (surface density rho)	double integral over R of rho dA
First moment about the y-axis, My	double integral over R of x dA
First moment about the x-axis, Mx	double integral over R of y dA
Centroid xbar	My / A = (1/A) double integral over R of x dA
Centroid ybar	Mx / A = (1/A) double integral over R of y dA

Worked example: Find the centroid of the region under y = x^2, above y = 0, from x = 0 to x = 2. The area is A = integral from 0 to 2 of x^2 dx = [x^3/3] from 0 to 2 = 8/3. The moment about the y-axis is

My = integral from 0 to 2 of integral from 0 to x^2 of x dy dx = integral from 0 to 2 of x(x^2) dx = integral from 0 to 2 of x^3 dx = [x^4/4] from 0 to 2 = 4,

so xbar = My / A = 4 / (8/3) = 3/2 = 1.5. The moment about the x-axis is Mx = integral from 0 to 2 of integral from 0 to x^2 of y dy dx = integral from 0 to 2 of (x^2)^2 / 2 dx = (1/2) integral from 0 to 2 of x^4 dx = (1/2)(32/5) = 16/5, so ybar = Mx / A = (16/5) / (8/3) = 6/5 = 1.2. The centroid sits at (1.5, 1.2), inside the region as it must.

Exam-day strategy

• To take a partial derivative, cover up every variable except the one you are differentiating and treat the rest as plain constants; the constant multiple just comes along for the ride.
• Use f_xy = f_yx as a free check on any second-partial problem: compute the mixed partial both ways, and if they disagree you have an arithmetic error to hunt down.
• For error propagation, prefer the relative form dz / z, where a product of powers becomes a weighted sum such as 2 dr / r + dh / h; the exponent of each variable is exactly its weight.
• Remember the gradient does two jobs at once: it points the way steepest uphill and its length is that steepest rate, while any direction perpendicular to it (along a level curve) gives a directional derivative of zero.
• Always normalize the direction vector to unit length before taking D_u f = grad f dot u; skipping that step inflates the answer by the length of the raw vector.
• Set the double integral up by drawing the region and a representative vertical strip first: the strip's ends give the inner (dy) limits as functions of x, and the leftmost and rightmost x-values give the outer (dx) limits.
• Pick the integrand by what is asked: dA for area, f(x, y) dA for volume under a surface, x dA and y dA for the moments, then divide each moment by the area to land the centroid.