Difficulty: Intermediate | Prerequisites: Partial derivatives, basic integration from MATH 124/125.
This covers two topics that appear together on MATH 126 Midterm 2: linear approximation (tangent planes) and double integrals. Linear approximation lets you estimate function values near a known point using partial derivatives. Double integrals extend single-variable integration to two dimensions, and reversing the order of integration is a standard technique for simplifying a computation. Both are fundamental tools in multivariable calculus and will reappear throughout the rest of the course.
The linear approximation of f(x, y) near a point (a, b) is a tangent-plane formula built from the value and both partial derivatives at that point. Double integrals over rectangles are computed as iterated integrals (integrate one variable at a time). When the region is not a rectangle, you may need to reverse the order of integration, which requires sketching the region and rewriting the bounds.
Linear approximation (linearisation)
The function L(x, y) = f(a, b) + f_x(a, b)(x − a) + f_y(a, b)(y − b). It gives the best linear estimate of f near the point (a, b). Think of it as the multivariable version of the tangent line approximation from Calc I.
Tangent plane
The plane z = L(x, y) that just touches the surface z = f(x, y) at the point (a, b, f(a, b)). The linear approximation and the tangent plane equation are the same formula.
Total differential
The quantity dz = f_x dx + f_y dy, which estimates the change in f when x and y change by small amounts dx and dy. Closely related to linear approximation: dz ≈ Δz for small changes.
Double integral
The integral ∬_R f(x, y) dA, which computes the signed volume under the surface z = f(x, y) over a region R in the xy-plane. In simple terms, it is the two-dimensional version of the area-under-a-curve integral.
Iterated integral
A double integral written as two nested single-variable integrals. For a rectangle, ∬_R f dA = ∫∫ f(x,y) dy dx (or dx dy). You integrate the inner variable first, treating the outer variable as a constant.
Order of integration
Whether you integrate with respect to x first (dx dy) or y first (dy dx). On a rectangle the order does not matter. On a non-rectangular region, one order may be much simpler than the other.
Fubini's Theorem
The theorem that lets you compute a double integral as an iterated integral, and that lets you swap the order of integration, provided the function is continuous on the region.
For f(x, y) near the point (a, b):
L(x, y) = f(a, b) + f_x(a, b) · (x − a) + f_y(a, b) · (y − b)
This is a plane that matches the surface's value and slope at (a, b). It gives good estimates when (x, y) is close to (a, b).
Evaluate f(a, b).
Compute f_x and f_y, then evaluate each at (a, b).
Plug everything into the formula above.
To approximate f at a nearby point, substitute the nearby coordinates into L.
This is Problem 3 from the mock midterm.
Step 1: Evaluate f(1, 4).
f(1, 4) = √(1 · 4) = √4 = 2.
Step 2: Compute partial derivatives.
f(x, y) = (xy)^(1/2).
f_x = (1/2)(xy)^(−1/2) · y = y / (2√(xy)).
f_y = (1/2)(xy)^(−1/2) · x = x / (2√(xy)).
Step 3: Evaluate at (1, 4).
f_x(1, 4) = 4 / (2√4) = 4/4 = 1.
f_y(1, 4) = 1 / (2√4) = 1/4.
Step 4: Write the linear approximation.
L(x, y) = 2 + 1 · (x − 1) + (1/4) · (y − 4).
Simplified: L(x, y) = x + y/4.
Step 5: Use it to approximate f(1.01, 3.98).
L(1.01, 3.98) = 1.01 + 3.98/4 = 1.01 + 0.995 = 2.005.
Rounded to two decimal places: 2.01.
For a rectangular region R = {(x, y) | a ≤ x ≤ b, c ≤ y ≤ d}:
∬_R f(x, y) dA = ∫_a^b ∫_c^d f(x, y) dy dx = ∫_c^d ∫_a^b f(x, y) dx dy.
Both orders give the same answer (Fubini's Theorem). Pick whichever is easier to integrate.
Choose an order (say, dy dx).
Integrate the inner integral with respect to y, treating x as a constant.
Plug in the y-limits.
Integrate the resulting expression with respect to x.
Plug in the x-limits.
This is Problem 4(a) from the mock midterm.
Integrate with respect to y first (dy dx).
∫_1^16 ∫_1^4 (x/y + y/x) dy dx.
Inner integral (holding x fixed):
∫_1^4 (x/y + y/x) dy = x · ln(y) + y²/(2x), evaluated from y = 1 to y = 4.
= x · ln(4) + 16/(2x) − x · ln(1) − 1/(2x)
= x · ln(4) + 8/x − 1/(2x)
= x · ln(4) + 15/(2x).
Now integrate with respect to x from 1 to 16.
∫_1^16 [x · ln(4) + 15/(2x)] dx
= ln(4) · x²/2 + (15/2) · ln(x), evaluated from x = 1 to x = 16.
= ln(4) · 256/2 + (15/2) · ln(16) − ln(4) · 1/2 − (15/2) · ln(1)
= 128 · ln(4) + (15/2) · ln(16) − (1/2) · ln(4) − 0
= (255/2) · ln(4) + (15/2) · ln(16).
Since ln(16) = 4 · ln(2) and ln(4) = 2 · ln(2):
= (255/2) · 2 · ln(2) + (15/2) · 4 · ln(2)
= 255 · ln(2) + 30 · ln(2)
= 285 · ln(2).
Sometimes the inner integral is impossible (or extremely hard) to evaluate in the given order. Switching the order of integration can turn an impossible integral into a routine one.
Sketch the region of integration from the original limits.
Describe the same region with the variables swapped: if the original is dy dx, rewrite as dx dy (or vice versa).
Determine the new outer limits (constants) and new inner limits (functions of the outer variable).
Write the new iterated integral.
This is Problem 4(b) from the mock midterm.
Step 1: Identify the region from the original bounds.
The outer integral says 0 ≤ y ≤ 3. The inner integral says 3y ≤ x ≤ 9.
So the region is: 0 ≤ y ≤ 3, 3y ≤ x ≤ 9.
Step 2: Sketch the region.
The boundary lines are x = 3y (i.e. y = x/3), x = 9, y = 0, and y = 3. The region is a triangle with vertices at (0, 0), (9, 0), and (9, 3).
Step 3: Rewrite the bounds with x as the outer variable.
x ranges from 0 to 9. For a fixed x, y ranges from 0 to x/3 (since y = x/3 is the boundary line solved for y).
Step 4: Write the reversed integral.
∫_0^9 ∫_0^{x/3} 11e^(x²) dy dx.
Note: you are not asked to evaluate this integral. The point is that the inner integral with respect to y is now trivial (11e^(x²) is a constant with respect to y), which would make the computation possible if you needed to continue.
L(x,y) = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)This is the linear approximation (tangent plane) of f at the point (a, b).
\iint_R f(x,y)\,dA = \int_a^b \int_c^d f(x,y)\,dy\,dxIterated integral over the rectangle [a, b] × [c, d]. Integrate inner variable first.
Linear approximation is used constantly in physics and engineering for small-perturbation analysis. If you know a system's behaviour at one operating point, the linearisation tells you how it responds to small changes, which is the basis of control theory and error analysis.
Double integrals compute quantities like mass, charge, probability, and average value over two-dimensional regions. If you have a density function over a plate or membrane, the double integral gives you the total mass.
Students often confuse the linear approximation formula with the equation of the tangent plane in the form z − z₀ = f_x(x − a) + f_y(y − b). They are the same thing, just rearranged. Use whichever the problem expects.
When computing partial derivatives of something like √(xy), students sometimes forget the chain rule and drop the inner derivative. Remember: d/dx of √(xy) needs the partial of (xy) with respect to x, which is y.
In double integrals, students treat the inner variable's limits as constants when they should be functions (or vice versa). On a rectangle, all limits are constants. On a non-rectangular region, the inner limits are functions of the outer variable.
When reversing the order of integration, students often skip sketching the region. This almost always leads to wrong bounds. Draw it.
⚠️ Linear approximation is a reliable 8-9 point question on Midterm 2. Know the formula, be able to compute partial derivatives of common functions (square roots, products, quotients), and be ready to use L to approximate a nearby value.
⚠️ Double integrals over rectangles: you must give answers in exact form. Leave answers as multiples of ln(2) or similar, not as decimals.
⚠️ Reversing the order of integration: always sketch the region. The exam often gives an integral whose inner function has no elementary antiderivative in the given order (like e^(x²)), making reversal the entire point of the problem.
⚠️ Watch your algebra when evaluating iterated integrals. Errors in plugging in limits are the most common source of lost marks on these problems.
True or False: The linear approximation of f at (a, b) requires only f(a, b) and the two partial derivatives at (a, b).
True.
Fill in the blank: In an iterated integral, you integrate the ______ variable first, treating the ______ variable as a constant.
Inner; outer.
True or False: You can always swap the order of integration in a double integral without changing the limits.
False. On a non-rectangular region, you must re-derive the limits for the new order.
Fill in the blank: Reversing the order of integration for ∫0^3 ∫{3y}^9 f(x, y) dx dy gives outer limits x = ______ to x = ______ and inner limits y = ______ to y = ______.
x = 0 to x = 9; y = 0 to y = x/3.
Q: Find the linear approximation of f(x, y) = e^(x+y) at the point (0, 0).
A: f(0, 0) = e^0 = 1. f_x = e^(x+y), so f_x(0, 0) = 1. f_y = e^(x+y), so f_y(0, 0) = 1. L(x, y) = 1 + 1 · (x − 0) + 1 · (y − 0) = 1 + x + y.
Q: Use the linear approximation from the worked example to estimate √(1.02 · 3.96).
A: L(x, y) = x + y/4. L(1.02, 3.96) = 1.02 + 3.96/4 = 1.02 + 0.99 = 2.01.
Q: Compute ∫_0^2 ∫_0^1 (x + 2y) dx dy.
A: Inner integral: ∫_0^1 (x + 2y) dx = [x²/2 + 2xy] from 0 to 1 = 1/2 + 2y. Outer integral: ∫_0^2 (1/2 + 2y) dy = [y/2 + y²] from 0 to 2 = 1 + 4 = 5.
Q: Reverse the order of integration for ∫_0^1 ∫_0^(√x) f(x, y) dy dx.
A: Region: 0 ≤ x ≤ 1, 0 ≤ y ≤ √x. Equivalently, y² ≤ x ≤ 1, 0 ≤ y ≤ 1. Reversed: ∫0^1 ∫(y²)^1 f(x, y) dx dy.
Linear approximation connects to the gradient and directional derivatives: the gradient vector ∇f = (f_x, f_y) encodes all the slope information the linearisation uses. It also connects to differentials and error propagation, which appear in physics and engineering courses.
Double integrals are the foundation for triple integrals, surface integrals, and the change-of-variables formula (polar, cylindrical, spherical coordinates), all of which come later in MATH 126. Reversing the order of integration is a special case of the more general idea of choosing the right coordinate system for a problem.
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