McGill University · MATH 222

McGill MATH 222: Calculus 3 tutoring

Work with an online tutor on partial derivatives, optimization and multiple integrals in McGill MATH 222. You can start with a question or a concept you want explained. Your tutor can help you sketch the surface or region, identify what stays fixed in a derivative and build integration bounds from the picture. If you have calculations, you can check them together.

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Course overview

McGill MATH 222 develops calculus in several variables through vector geometry, partial derivatives, optimization, coordinates, and multiple integrals. It also includes Taylor series in one and several variables.

MATH 222 topics to review

Taylor seriesVector geometryPartial and directional derivativesMultivariable optimizationParametric curves and arc lengthMultiple integrals and coordinates

Common difficulties

Translating a region into correct multiple-integral bounds
Distinguishing partial, directional and total changes

Choose the obstacle, then the example

My region does not match my bounds. Use the existing triangle and change-of-order example.
I found a stationary point but cannot classify it. Compare the minimum, saddle and inconclusive test already explained below.
I have a polynomial but no accuracy guarantee. Use this single-variable Taylor refresher before returning to the multivariable work in your outline.

Preparing for this course

Combine single-variable calculus with vector geometry and sketches. Treat the variables held fixed explicitly when working with partial derivatives.

Registration requirements

The guidance above is study preparation, not an eligibility decision. Confirm current prerequisites, restrictions and availability with McGill University.

Check the official course catalogue

Try a course problem

Let $x,y$ be independent real inputs and $f(x,y)=x^2+3xy$ . Find the partial derivative $\partial f/\partial x$ .

Show the reasoning

A partial derivative with respect to $x$ holds $y$ constant. The term $x^2$ contributes $2x$ ; in $3yx$ , the constant coefficient is $3y$ , so its derivative is $3y$ . Therefore $\partial f/\partial x=2x+3y$ .

Let the region determine the integration bounds

A double integral describes a region as well as an integrand. This example shows how a sketch determines dependent bounds and why reversing the order must preserve the same triangle.

1. Identify the boundary lines and vertices

Let $x$ and $y$ be horizontal and vertical coordinates in the plane. Let $D$ name the region of integration, $I$ the integral’s value, and $dA$ an infinitesimal element of area. Let $D$ satisfy $x\ge0$ , $y\ge0$ and $x+y\le2$ . Its vertices are $(0,0)$ , $(2,0)$ and $(0,2)$ . Evaluate $I=\iint_D(x+2y)\,dA.$ The sloping boundary is $y=2-x$ , so the available height changes with $x$ .

The triangle is bounded by the coordinate axes and

x+y=2

. At fixed

x

, a vertical slice runs from

y=0

y=2-x

2. Use vertical slices

For each fixed $x$ between $0$ and $2$ , the coordinate $y$ runs from $0$ to $2-x$ . Hence $I=\int_0^2\int_0^{2-x}(x+2y)\,dy\,dx.$ Treat $x$ as constant in the inner integral: $\begin{aligned}&\int_0^{2-x}(x+2y)\,dy\\&=x(2-x)+(2-x)^2\\&=4-2x.\end{aligned}$ Therefore $I=\int_0^2(4-2x)\,dx=4$ .

3. Reverse the order by describing horizontal slices

For fixed $y$ , the horizontal coordinate runs from $0$ to $2-y$ , while $0\le y\le2$ . Thus $I=\int_0^2\int_0^{2-y}(x+2y)\,dx\,dy.$ The inner integral becomes $2+2y-\tfrac32y^2$ ; integrating from $0$ to $2$ again gives $4$ . The order changed, but neither the region nor the integrand changed.

4. Check geometry as well as algebra

The triangle’s area is $2$ . The average value of $x+2y$ over it is therefore $4/2=2$ , which lies between its minimum $0$ and maximum $4$ on the triangle. This is a useful check, though it does not by itself prove the answer.

Try it yourself

For the triangle $x\ge0$ , $y\ge0$ , $x+2y\le4$ , write its area as an iterated integral (one integral inside another) in both orders and evaluate it.

Show the explained solution

Let $A$ denote the requested area. Integrating 1 adds the area elements without weighting them. Vertical slices give $\begin{gathered}A\\=\int_0^4\int_0^{(4-x)/2}1\,dy\,dx\\=4.\end{gathered}$ Horizontal slices give $\begin{gathered}A\\=\int_0^2\int_0^{4-2y}1\,dx\,dy\\=4.\end{gathered}$ The base is $4$ and the height is $2$ , agreeing with the triangle area formula.

Before calculating, describe one slice in words and test a point inside the proposed bounds. A correct antiderivative cannot repair a region that was described incorrectly.

Stationary points: finding a candidate is only the first step

Let $x$ and $y$ be independent real variables and consider $\begin{aligned}f(x,y)&=x^2+xy+y^2\\&\quad-4x-2y.\end{aligned}$ Find and classify its stationary point. Is your conclusion local or global?

Classify the point, then prove the global result

First set both partial derivatives to zero: $\begin{aligned}f_x&=2x+y-4=0,\\f_y&=x+2y-2=0.\end{aligned}$ Solving gives $(2,0)$ , where $f=-4$ . Zero first derivatives alone do not establish a minimum.

For continuous second partial derivatives near a stationary point, define $D=f_{xx}f_{yy}-(f_{xy})^2.$ Here $f_{xx}=f_{yy}=2$ and $f_{xy}=1$ , so $D=3>0$ . Together with $f_{xx}>0$ , this establishes a strict local minimum. The test gives a strict local maximum if $D>0$ and $f_{xx}<0$ , a saddle if $D<0$ , and no conclusion when $D=0$ .

For this function, completing the square proves more: $\begin{aligned}f(x,y)+4&=\left(x-2+\frac y2\right)^2\\&\quad+\frac{3y^2}{4}.\end{aligned}$ Both terms are nonnegative and vanish together only at $(2,0)$ . Thus $-4$ is the unique global minimum value.

Compare a saddle and an inconclusive test

Compare $g(x,y)=x^2-y^2$ at the origin with $h(x,y)=(x+y)^2$ on the line $x+y=0$ . Does an inconclusive second-derivative test rule out a minimum?

Check your reasoning

For $g$ , the gradient is zero at the origin but $D=-4$ . The function takes positive and negative values arbitrarily nearby: a saddle.

For $h$ , we have $D=0$ and the test is inconclusive, but $h\ge0$ and it equals zero on $x+y=0$ . Every point on that line is a non-strict global minimum. Inconclusive does not mean no minimum exists.

For a problem restricted to a region, inspect its boundary as well as interior stationary points.

For the derivative interpretation, review gradients and directional derivatives.

For related single-variable review, practise Taylor approximation and error bounds.

How tutoring can help

Your tutor can connect your partial-derivative or gradient calculation to a surface sketch, clarifying what is held fixed and which direction is measured. For multiple integrals, you can examine your region and proposed bounds together before practising a change of order or coordinates.

Exam preparation for MATH 222

Check that bounds describe the intended region exactly once. Practise coordinate changes, directional derivatives and optimization with explicit constraints from your outline.

We work through permitted practice and comparable problems, without completing graded assignments or exams on your behalf.

Independent practice

Free practice and study guides for MATH 222

6 questions

Double integrals: regions and changing order

Practise double-integral bounds, curved regions, splitting a region, reversing integration order and polar coordinates, with diagrams and explained solutions.

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