On changing the order of integration, ∫₀¹ ∫_y¹ e^{x²} dx dy equals:

The integral cannot be evaluated in closed form

At a point P, a function f(x, y) has fₓ = f_y = 0 and D = fₓₓ·f_yy − (fₓ_y)² = 0. Which of the following must be true? (More than one option may be correct.)

The second-derivative test yields no conclusion at P

P is necessarily a saddle point

P must be a point of local minimum

Multivariable Calculus — Partial Derivatives, Maxima and Multiple Integrals

Single-variable calculus asks how a function changes; multivariable calculus asks how it changes in which direction, and almost every difficulty in this chapter comes from forgetting that the direction has to be specified. A partial derivative is nothing more mysterious than an ordinary derivative taken while the other variables are held still — differentiate with respect to x and treat y as a constant, no differently from the way you treat 5. Two habits pay for themselves immediately. First, in a double integral the limits belong to the variable that is integrated first, and reading them off the wrong way round is the single largest source of lost marks in the topic; sketching the region takes fifteen seconds and settles it permanently. Second, the second-derivative test for two variables is not two separate tests. You compute one number, D = fₓₓ·f_yy − (fₓ_y)², and only its sign matters first: D < 0 is a saddle point regardless of anything else, D > 0 sends you to the sign of fₓₓ for minimum or maximum, and D = 0 tells you nothing at all and the test has failed. Candidates lose marks by checking fₓₓ before D, which answers a question the test has not yet reached. Everything else here is mechanical: change the order of integration when the inner integral has no elementary antiderivative, switch to polar coordinates the moment you see x² + y² in the integrand or a circular region, and remember that the Jacobian factor r in r dr dθ is part of the substitution rather than an optional flourish.

Partial Derivatives, the Total Derivative and Euler’s Theorem

A partial derivative freezes the other variables. For f = x³ + x²y + y³, fₓ = 3x² + 2xy and f_y = x² + 3y². Nothing about the procedure is new; only the bookkeeping is.
Mixed partials commute for the functions GATE sets: fₓ_y = f_yₓ. So a question that asks for one and gives you the other has already been answered, and computing the harder order is wasted work.
The total derivative chains through every path. If z = f(x, y) with x and y both functions of t, then dz/dt = fₓ·(dx/dt) + f_y·(dy/dt). For z = x²y with x = t² and y = t³ this gives 4t⁶ + 3t⁶ = 7t⁶, so dz/dt = 7 at t = 1 — and substituting first gives z = t⁷, whose derivative is 7t⁶. The two routes must agree, which makes substitution a free check.
Euler’s theorem is a free identity for homogeneous functions. If every term of f has the same total degree n, then x·fₓ + y·f_y = n·f. For f = x³ + x²y + y³, every term has degree 3, so the combination is exactly 3f — verified term by term. Questions that look like they need a page of differentiation often collapse to this in one line.

🧠 Mean value theorems and indeterminate forms, in the form GATE uses them

Rolle needs f continuous on [a, b], differentiable on (a, b) and f(a) = f(b); Lagrange drops the last condition and concludes f′(c) = [f(b) − f(a)]/(b − a). GATE almost never asks you to find c — it asks whether the hypotheses hold, and the answer is usually no because the function has a corner or a vertical tangent somewhere inside. For indeterminate forms, apply L’Hôpital only to 0/0 and ∞/∞, and re-check the form after every application: lim(x→0) (x − sin x)/x³ needs the rule three times and comes out to 1/6, and stopping one step early gives 0/0 rather than an answer.

Maxima, Minima and Saddle Points

Set fₓ = 0 and f_y = 0 and solve the pair — those are the critical points, and there is usually more than one. Then evaluate D = fₓₓ·f_yy − (fₓ_y)² at each one separately. A single function can carry a minimum at one critical point and a saddle at another, so a question that gives you a function and one point is testing one evaluation, not the whole landscape.

The second-derivative test, worked on f = x³ + y³ − 3xy
Critical point	D = fₓₓ·f_yy − (fₓ_y)²	Verdict
(0, 0)	0·0 − (−3)² = −9	D < 0 → saddle point. The sign of fₓₓ is irrelevant here and must not be consulted.
(1, 1)	6·6 − (−3)² = 27	D > 0 and fₓₓ = 6 > 0 → local minimum.

⚠️ D = 0 is not "neither" — it is "the test failed"

When D = 0 the test is silent, and the point may still be a maximum, a minimum or a saddle. If an option list offers "no conclusion can be drawn from the second-derivative test", that is a real answer and often the correct one. Treating D = 0 as proof of a saddle is a definite error, and it is a favourite distractor precisely because it feels like a verdict.

Double and Triple Integrals, and When to Change Something

Read the limits inside out. In ∫₀¹ ∫₀ˣ (x + y) dy dx the inner integral runs over y from 0 to x with x held fixed, giving xy + y²/2 evaluated at y = x, that is 3x²/2; the outer integral then gives 1/2. If you had read the x-limits as the inner ones you would have integrated over a different region and got a different number.
Change the order when the inner integral cannot be done. ∫₀¹ ∫_y¹ e^x² dx dy is hopeless as written, because e^x² has no elementary antiderivative. Sketch the region — it is the triangle 0 ≤ y ≤ x ≤ 1 — and rewrite it as ∫₀¹ ∫₀ˣ e^x² dy dx. Now the inner integral is trivial, the integrand becomes x·e^x², and the answer is (e − 1)/2. An integral that looks impossible is a signal to change the order, not to move on.
Go polar on sight of x² + y². Over the unit disc, ∫∫ (x² + y²) dA becomes ∫₀^2π ∫₀¹ r²·r dr dθ = 2π·(1/4) = π/2. The extra r is the Jacobian; omitting it is the standard error, and it turns an area element into a length element.
A triple integral over a box separates into a product. ∫∫∫ xyz over the unit cube is (∫₀¹ x dx)(∫₀¹ y dy)(∫₀¹ z dz) = (1/2)³ = 1/8. Separation works only when both the integrand factorises and the limits are constants — the moment a limit mentions another variable, the shortcut is gone.

Key takeaways

A partial derivative is an ordinary derivative with the other variables held constant. Mixed partials commute, so compute whichever order is easier.
The total derivative adds one term per path: dz/dt = fₓ·dx/dt + f_y·dy/dt. Substituting first is an independent check on the same answer.
For a function homogeneous of degree n, x·fₓ + y·f_y = n·f. Spotting the common degree replaces a page of differentiation.
Evaluate D = fₓₓ·f_yy − (fₓ_y)² first. D < 0 is a saddle whatever else is true; D > 0 sends you to fₓₓ; D = 0 means the test failed, not that the point is neither.
In a double integral the innermost limits belong to the variable integrated first. Sketching the region prevents the commonest error in the topic.
An inner integral with no elementary antiderivative, such as e^x², is an instruction to change the order of integration — not a dead end.
Switch to polar coordinates when you see x² + y² or a circular region, and carry the Jacobian: dA = r dr dθ.

Practice questions (9)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

For f(x, y) = x³ + y³ − 3xy, the point (0, 0) is:
1. A point of local minimum
2. A point of local maximum
3. A saddle point
4. Not a critical point at all
Show answer
Answer: C — A saddle point
fₓ = 3x² − 3y and f_y = 3y² − 3x both vanish at (0, 0), so it is a critical point and option D is out. Now fₓₓ = 6x, f_yy = 6y and fₓ_y = −3, so at the origin D = 0·0 − (−3)² = −9 < 0, which makes it a saddle point — and the sign of fₓₓ must not be consulted, because it decides between maximum and minimum only once D is positive. The same function has a genuine local minimum at its other critical point (1, 1), where D = 36 − 9 = 27 > 0 and fₓₓ = 6 > 0, which is why a question about this function has to name the point.
The value of ∫₀¹ ∫₀ˣ (x + y) dy dx is:
1. 1/2
2. 1/3
3. 1
4. 3/2
Show answer
Answer: A — 1/2
Integrate over y first, treating x as a constant: ∫₀ˣ (x + y) dy = [xy + y²/2] from 0 to x = x² + x²/2 = 3x²/2. Then ∫₀¹ (3x²/2) dx = (3/2)·(1/3) = 1/2. The region is the triangle below the line y = x in the unit square, and it is worth sketching once: if you mistakenly integrate x from 0 to y and y from 0 to 1 you are working over the other triangle, which for this integrand gives 1/2 as well by symmetry — a coincidence that hides the error here and will not hide it in the next question. Read the limits inside out every time.
On changing the order of integration, ∫₀¹ ∫_y¹ e^x² dx dy equals:
1. (e − 1)/2
2. e − 1
3. e/2
4. The integral cannot be evaluated in closed form
Show answer
Answer: A — (e − 1)/2
As written the inner integral asks for the antiderivative of e^x², which has none in elementary terms — and that is the hint, not the obstacle. The region is 0 ≤ y ≤ x ≤ 1, the triangle below y = x, so swapping the order gives ∫₀¹ ∫₀ˣ e^x² dy dx. The inner integral is now just e^x²·x, and ∫₀¹ x·e^x² dx = [e^x²/2] from 0 to 1 = (e − 1)/2. Option D is the answer a candidate reaches by taking the impossibility of the inner integral at face value, and it is wrong for the whole family of questions built this way: an inner integrand with no antiderivative is the examiner telling you to change the order.
For f(x, y) = x³ + x²y + y³, the expression x·∂f/∂x + y·∂f/∂y equals:
1. f
2. 2f
3. 3f
4. 0
Show answer
Answer: C — 3f
Every term of f has total degree 3 — x³, x²y and y³ all add to 3 in their exponents — so f is homogeneous of degree 3 and Euler’s theorem gives x·fₓ + y·f_y = 3f. Doing it the long way confirms it: x(3x² + 2xy) + y(x² + 3y²) = 3x³ + 2x²y + x²y + 3y³ = 3(x³ + x²y + y³). The examinable skill is recognising the common degree at a glance; the arithmetic above is the check, not the method. Note that a function like x³ + y² is not homogeneous, and Euler’s theorem simply does not apply to it.
The value of ∫∫ (x² + y²) dA over the disc x² + y² ≤ 1 is:
1. π/4
2. π/2
3. π
4. 2π
Show answer
Answer: B — π/2
In polar coordinates the integrand x² + y² is simply r², the region is 0 ≤ r ≤ 1 with θ from 0 to 2π, and the area element is r dr dθ. So the integral is ∫₀^2π ∫₀¹ r²·r dr dθ = 2π·[r⁴/4]₀¹ = 2π·(1/4) = π/2. Option A is what you get by dropping the Jacobian r and integrating r² dr dθ instead, which is the standard error in every polar question — the r is part of dA, not an optional extra. Option D is the area of the disc doubled and has no route to it; recognising that the answer must be smaller than the area π times the maximum integrand 1 is a useful sanity bound.
If z = x²y with x = t² and y = t³, then dz/dt at t = 1 is:
1. 5
2. 6
3. 7
4. 12
Show answer
Answer: C — 7
By the chain rule dz/dt = fₓ·(dx/dt) + f_y·(dy/dt) = 2xy·2t + x²·3t². Substituting x = t² and y = t³ gives 2t⁵·2t + t⁴·3t² = 4t⁶ + 3t⁶ = 7t⁶, which is 7 at t = 1. Substituting first is a free check and must agree: z = (t²)²·t³ = t⁷, so dz/dt = 7t⁶. Option A comes from adding the exponents of the two paths rather than their contributions, and option D from differentiating only the x-path and doubling — both are versions of the same mistake, dropping one term of the sum.
The value of lim(x→0) (x − sin x)/x³ is:
1. 0
2. 1/6
3. 1/3
4. The limit does not exist
Show answer
Answer: B — 1/6
The form is 0/0, so L’Hôpital applies — but it has to be applied until the form clears. Once: (1 − cos x)/3x², still 0/0. Twice: sin x/6x, still 0/0. Three times: cos x/6 → 1/6. The series route is faster and less error-prone: sin x = x − x³/6 + …, so x − sin x = x³/6 − … and dividing by x³ leaves 1/6. Option A is what you get by stopping after one application and reading 0/0 as 0, which is the commonest way this question is lost; option D ignores that the limit is a perfectly ordinary finite number.
The value of ∫₀¹ ∫₀¹ ∫₀¹ xyz dx dy dz is _____ (round to three decimal places if needed).
Numerical answer — type the value.
Show answer
Answer: 0.125
The integrand factorises and all six limits are constants, so the triple integral separates into a product of three identical single integrals: (∫₀¹ x dx)(∫₀¹ y dy)(∫₀¹ z dz) = (1/2)(1/2)(1/2) = 1/8 = 0.125. Separation is only legitimate under both conditions — a factorising integrand and constant limits. Change the upper limit of x to y and the shortcut vanishes, because the x-integral then depends on y and cannot be pulled out of the product. As a numerical-answer question there is no penalty for a wrong entry, so type the value rather than leaving it blank.
At a point P, a function f(x, y) has fₓ = f_y = 0 and D = fₓₓ·f_yy − (fₓ_y)² = 0. Which of the following must be true? (More than one option may be correct.)
1. The second-derivative test yields no conclusion at P
2. P is necessarily a saddle point
3. P is not a critical point of f
4. P must be a point of local minimum
Show answer
Answer: A — The second-derivative test yields no conclusion at P
Only option A is true. D = 0 is the inconclusive case: the point may be a maximum, a minimum or a saddle, and the test cannot tell you which — so B and D both claim more than the data allows. Option C is simply false, since fₓ = f_y = 0 is the definition of a critical point. Note what this question is really testing, and it is not calculus: an MSQ may have exactly one correct option, and a candidate who reads "more than one option may be correct" as "at least two are correct" will pair A with B or D and score nothing, because GATE gives no partial credit. Select what you can justify and nothing else.