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Multivariable Calculus: free practice, theory and problems

A function of several variables, such as , describes something that depends on more than one quantity – temperature in a room, cost as a function of two design parameters, or pressure as a function of position. The partial derivatives tell you how fast changes as you move along one axis at a time, and the gradient collects them into a vector that points in the direction where increases fastest. This is the foundation of optimization with several variables: finding the cheapest design, the strongest structure, or the hottest spot on a plate.

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Partial derivatives
Multiple integrals
Vector calculus

1. Partial derivatives

What is it about?

A function of several variables, such as $f(x, y)$ , describes something that depends on more than one quantity – temperature in a room, cost as a function of two design parameters, or pressure as a function of position. The partial derivatives tell you how fast $f$ changes as you move along one axis at a time, and the gradient $\nabla f$ collects them into a vector that points in the direction where $f$ increases fastest. This is the foundation of optimization with several variables: finding the cheapest design, the strongest structure, or the hottest spot on a plate.

Key concepts and formulas

Partial derivative $f_x = \partial f/\partial x$ : differentiate with respect to $x$ , treating $y$ as a constant (and vice versa for $f_y$ ).
Gradient: $\nabla f = (f_x, f_y)$ . It points in the direction where $f$ increases fastest, and is perpendicular to the level curves.
Directional derivative along the unit vector $\vec u$ : $D_{\vec u}f = \nabla f\cdot\vec u$ .
Chain rule with one parameter: $\dfrac{dz}{dt} = f_x\dfrac{dx}{dt} + f_y\dfrac{dy}{dt}$ . With two parameters $u, v$ : $\dfrac{\partial z}{\partial u} = f_x\dfrac{\partial x}{\partial u} + f_y\dfrac{\partial y}{\partial u}$ .
Critical point: where $\nabla f = \vec 0$ (or where the partial derivatives fail to exist).
Second derivative test: $D = f_{xx}f_{yy} - f_{xy}^2$ . $D>0, f_{xx}>0$ : local minimum. $D>0, f_{xx}<0$ : local maximum. $D<0$ : saddle point.
Lagrange's method: an extremum of $f$ subject to the constraint $g=0$ satisfies $\nabla f = \lambda\nabla g$ together with $g=0$ .
Tangent plane at $(a,b)$ : $z = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)$ . The same expression gives the linear approximation.

How to solve the problems

Differentiate partially: hold every other variable constant.
For critical points: set $f_x=0$ and $f_y=0$ , and solve the system of equations.
Classify with the second derivative test, or use Lagrange's method if there is a constraint.
For a directional derivative: normalize the direction vector first (divide by its length) if it does not already have length 1.
For a linear approximation or tangent plane: compute $f$ and the gradient at the point, and substitute into the formula.

Example

Find and classify the critical points of $f(x,y) = x^2 + y^2 - 2x - 4y + 5$ .

$f_x = 2x - 2 = 0 \Rightarrow x=1$ . $f_y = 2y-4=0 \Rightarrow y=2$ . Critical point: $(1,2)$ .
$f_{xx}=2$ , $f_{yy}=2$ , $f_{xy}=0$ , so $D = 2\cdot 2 - 0^2 = 4 > 0$ and $f_{xx}>0$ .
The point $(1,2)$ is a local minimum, with $f(1,2) = 1+4-2-8+5 = 0$ .

Common mistakes

Forgetting to normalize the direction vector before using $D_{\vec u}f = \nabla f\cdot\vec u$ .
Assuming $f_{xy}$ and $f_{yx}$ could differ – for smooth functions they are equal (Clairaut's theorem), but still write out the computation carefully.
Believing $\nabla f = \vec 0$ always means an extremum. It can also be a saddle point.
Writing the Lagrange condition backwards: it is $\nabla f = \lambda\nabla g$ , not $\nabla g = \lambda\nabla f$ .

The gradient points where

f

increases fastest. A critical point is where

\nabla f = \vec 0

– the second derivative test decides whether it is a minimum, a maximum or a saddle point.

Concepts in this part

Partial derivative and gradientA partial derivative differentiates with respect to one variable at a time, holding the others constant. The g…The second derivative testAt a critical point where , the second derivative test decides whether it is a local minimum, maximum or a sad…Lagrange multipliersLagrange's method finds extreme points of when is constrained to a curve . At the extreme point the level curv…

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2. Multiple integrals

What is it about?

A single integral sums values along a line. A double integral $\iint_D f\,dA$ sums values over a region in the plane, and a triple integral $\iiint_B f\,dV$ sums over a volume. They are used for area, volume, mass, centers of mass and average values – anything that is "the sum of many small contributions" over a surface or a volume. An engineer uses this to, for example, find the mass of a plate with varying density, or the volume of a tank with slanted walls.

Key concepts and formulas

Double integral over a rectangle: $\iint_{[a,b]\times[c,d]} f(x,y)\,dA = \int_a^b\!\int_c^d f(x,y)\,dy\,dx$ (Fubini's theorem: the order can be swapped for continuous functions).
Area: $\iint_D 1\,dA$ . Volume under a surface: $\iint_D f\,dA$ when $f\ge 0$ .
Mass with density $\rho(x,y)$ : $m = \iint_D \rho\,dA$ .
Polar coordinates: $x=r\cos\theta$ , $y=r\sin\theta$ , $dA = r\,dr\,d\theta$ . The factor $r$ is the Jacobian.
Cylindrical coordinates: $dV = r\,dr\,d\theta\,dz$ . Spherical coordinates: $dV = \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$ , where $\varphi$ is measured from the $z$ -axis.
Triple integral over a box: $\iiint_{[a,b]\times[c,d]\times[e,f]} 1\,dV = (b-a)(d-c)(f-e)$ .

How to solve the problems

Describe the region, and decide whether Cartesian, polar, cylindrical or spherical coordinates are simplest.
Set up the bounds from the inside out: the innermost variable can have bounds that depend on the outer ones.
Switch to polar/cylindrical/spherical by replacing $x,y,(z)$ and adding the correct Jacobian factor ( $r$ or $\rho^2\sin\varphi$ ).
Evaluate the integral from the inside out, one step at a time.
Check the result: is it positive where it should be (area, volume, mass)? Does it have the right order of magnitude?

Example

Find the area of the region between $y=x^2$ and $y=4$ .

The curves meet where $x^2=4$ , i.e. $x=\pm 2$ .
For each $x$ between $-2$ and $2$ , $y$ runs from $x^2$ to $4$ : $A = \int_{-2}^{2}\int_{x^2}^{4} dy\,dx = \int_{-2}^{2}(4-x^2)\,dx$ .
$\int_{-2}^2(4-x^2)\,dx = \left[4x - \tfrac{x^3}{3}\right]_{-2}^2 = \left(8-\tfrac83\right) - \left(-8+\tfrac83\right) = 16 - \tfrac{16}{3} = \tfrac{32}{3} \approx 10.67$ .

Common mistakes

Forgetting the factor $r$ in polar/cylindrical coordinates, or $\rho^2\sin\varphi$ in spherical coordinates.
Using fixed bounds when the region is not actually a rectangle or a box – the inner bound often depends on the outer variable.
Swapping the order of integration without updating the bounds accordingly.
Confusing $\varphi$ (measured from the $z$ -axis) with the polar angle $\theta$ in spherical coordinates.

Choose the coordinate system to match the shape of the region: rectangle/box → Cartesian, circle/disk → polar, cylinder → cylindrical, sphere → spherical.

Concepts in this part

Double integralA double integral sums a function over an area, and is usually computed as two ordinary integrals in a row. Fu…Polar coordinatesIn polar coordinates a point is described by its distance from the origin and angle . The area element gets an…Spherical coordinatesSpherical coordinates describe a point by its distance from the origin, a polar angle from the z-axis, and an …

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3. Vector calculus

What is it about?

A vector field $\vec F(x,y,z)$ describes something that has both direction and magnitude at every point – a flow velocity, a force field, a heat flux. Vector calculus gives tools to measure how much a field "spreads out" (divergence) and how much it "rotates" (curl), and the big theorems (Green, Stokes, the divergence theorem) tie these local measures to global quantities such as flux through a surface or circulation around a curve. This is the language of fluid mechanics, electromagnetism and heat transfer.

Key concepts and formulas

Divergence: $\nabla\cdot\vec F = \partial_xF_1+\partial_yF_2+\partial_zF_3$ . It measures sources/sinks – positive divergence means the field "flows out" of the point.
Curl: $\nabla\times\vec F$ . It measures rotation in the field. $\nabla\times(\nabla f)=\vec 0$ always (a gradient field is irrotational), and $\nabla\cdot(\nabla\times\vec F)=0$ always.
Conservative field: $\vec F=\nabla f$ for a potential function $f$ , on a simply connected region equivalent to $\nabla\times\vec F=\vec 0$ . Then $\int_C\vec F\cdot d\vec r = f(\text{end})-f(\text{start})$ , independent of the path.
Green's theorem (in the plane): $\oint_C P\,dx+Q\,dy = \iint_D(Q_x-P_y)\,dA$ .
Divergence theorem (Gauss): $\oiint_S\vec F\cdot d\vec S = \iiint_V\nabla\cdot\vec F\,dV$ .
Stokes' theorem: $\oint_C\vec F\cdot d\vec r = \iint_S(\nabla\times\vec F)\cdot d\vec S$ , where $C$ bounds the surface $S$ .

How to solve the problems

Check whether the field is conservative: is $\nabla\times\vec F=\vec 0$ ? If so, find the potential function and use only the endpoints.
For closed curves in the plane: use Green's theorem instead of computing the line integral directly.
For flux through a closed surface: use the divergence theorem, compute $\nabla\cdot\vec F$ and integrate over the volume.
For circulation around a curve that bounds a surface: use Stokes' theorem.
Check the sign and units at the end – flux and circulation can be negative.

Example

Use the divergence theorem to find the flux of $\vec F=(x^2,\,0,\,0)$ out of the cube $[0,2]^3$ .

$\nabla\cdot\vec F = 2x$ .
The flux is $\iiint_V 2x\,dV = \int_0^2\!\int_0^2\!\int_0^2 2x\,dx\,dy\,dz$ .
$\int_0^2 2x\,dx = 4$ , and the rest of the volume contributes a factor $2\cdot 2=4$ : the flux is $4\cdot 4=16$ .

Common mistakes

Using Green's/Stokes'/the divergence theorem on a surface or curve that is not closed.
Confusing divergence (a scalar) with curl (a vector) – divergence gives sources/sinks, curl gives rotation.
Forgetting that a conservative field makes the line integral path-independent – then you do not need to parametrize the curve at all.
Sign errors in Green's theorem: it is $Q_x-P_y$ , not $P_y-Q_x$ .

Divergence measures sources, curl measures rotation. The three big theorems trade a hard line/surface integral for an easier surface/volume integral.

Concepts in this part

Divergence and curlThe divergence of a vector field measures how much the field "flows out" from a point, while the curl measures…Green's theoremGreen's theorem relates a line integral around a closed curve to a double integral over the region inside. It …Divergence theorem and StokesThe divergence theorem (Gauss) relates the flux out of a closed surface to the volume integral of the divergen…

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Example problems with solutions

Here are some of the problems in multivariable Calculus. In the app, calculation problems get new numbers every time, so you can practise until it sticks – and take a graded practice exam before the real one.

Partial derivatives:

f(x, y) = x^2y

. What is

\partial f/\partial x

Answer: $2xy$

Treat $y$ as a constant.

Multiple integrals:

\displaystyle\iint_{[0,1]\times[0,1]} xy\,dA =

Answer: 0.25

$\int_0^1 x\,dx\cdot\int_0^1 y\,dy = \tfrac12\cdot\tfrac12 = \tfrac14 = 0.25$ .

Vector calculus:

\nabla\cdot(x, y, z) =

Answer: 3

$1 + 1 + 1 = 3$ .

Partial derivatives: The gradient

\nabla f

points in the direction where …

Answer: $f$ increases fastest

$\nabla f$ is perpendicular to the level curves.

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Matches these university courses

The content covers the syllabus found in engineering degrees, for example:

DAVE3700 (OsloMet)
TMA4420 (NTNU)
MATH112 (NMBU)

Multivariable Calculus: free practice, theory and problems

Contents

1. Partial derivatives

What is it about?

Key concepts and formulas

How to solve the problems

Example

Common mistakes

Concepts in this part

2. Multiple integrals

What is it about?

Key concepts and formulas

How to solve the problems

Example

Common mistakes

Concepts in this part

3. Vector calculus

What is it about?

Key concepts and formulas

How to solve the problems

Example

Common mistakes

Concepts in this part

Example problems with solutions

Matches these university courses