University Calculus III — Double Integrals Worksheet

11 of the 12 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one. 11 of the 12 questions are printed below. The other 1 is built on a diagram or a table of values that does not translate to the page, so it is in the free PDF — marked below where it would have come.

1. Q1Double Integrals over Rectangles and Iterated Integrals

   Let $R=[1,3]\times [0,2]$, that is $1\le x\le 3$ and $0\le y\le 2$, and let $f(x,y)=4xy+3y^2$.

   1. Evaluate ${\displaystyle {\iint }_{R}f(x,y)dA}$ as the iterated integral ${\displaystyle {\int }_1^3{\int }_0^{2}f(x,y)dydx}$.
   2. Evaluate it again in the order $dxdy$, and state the theorem that says the two values had to agree.
2. Q2Double Integrals over Rectangles and Iterated Integrals

   Evaluate $${\iint }_R\frac{x}{(1+xy{)}^2}dA,R=[0,2]\times [0,1].$$ One order of integration is much shorter than the other. Say which you chose and why.

3. Q3Double Integrals over General Regions

   This question is built around a diagram or a table of values. Open it in the PDF.

4. Q4Double Integrals over General Regions

   Let $D$ be the triangle with vertices $(0,0)$, $(3,3)$ and $(6,0)$.

   1. Describe $D$ by inequalities of the form $c\le y\le d$, $h_1(y)\le x\le h_2(y)$, and explain in one sentence why this description is more convenient here than one using vertical strips.
   2. Evaluate ${\displaystyle {\iint }_{D}xydA}$.
5. Q5Reversing the Order of Integration

   Consider ${\displaystyle {\int }_0^3{\int }_{x^2}^{3x}f(x,y)dydx}$.

   1. Sketch the region of integration and label its boundary curves and corner points.
   2. Rewrite the integral in the order $dxdy$.
   3. Check your limits by evaluating both iterated integrals for $f(x,y)=x$.
6. Q6Reversing the Order of Integration

   Evaluate $${\int }_0^2{\int }_{2y}^4{e}^{-x^2/4}dxdy.$$ The function $e^{-x^2/4}$ has no antiderivative that can be written with elementary functions, so begin by reversing the order of integration.

7. Q7Polar Coordinates and Polar Curves

   Polar coordinates $(r,\theta )$ are related to Cartesian ones by $x=r\cos \theta$, $y=r\sin \theta$, with $r\ge 0$.

   1. Find the polar equation of the circle $x^2+y^2=2y$, and give an interval of $\theta$, inside $0\le \theta \le 2\pi$, over which the circle is traced exactly once.
   2. Find the polar equation of the vertical line $x=2$.
   3. Let $D$ be the part of the disc $x^2+y^2\le 16$ lying to the right of the line $x=2$. Describe $D$ by inequalities of the form $\alpha \le \theta \le \beta$, $r_1(\theta )\le r\le r_2(\theta )$.
8. Q8Double Integrals in Polar Coordinates

   Let $D$ be the part of the ring $1\le x^2+y^2\le 4$ that lies in the first quadrant ($x\ge 0$, $y\ge 0$). Describe $D$ in polar coordinates and evaluate ${\displaystyle {\iint }_{D}xydA}$.

9. Q9Double Integrals in Polar Coordinates

   Sketch the region of integration of $${\int }_0^{\sqrt{2}}{\int }_x^{\sqrt{4-x^2}}\frac{1}{1+x^2+y^2}dydx,$$ convert the integral to polar coordinates, and evaluate it.

10. Q10Area and Volume by Double Integration

    Let $D$ be the triangle in the $xy$-plane with vertices $(0,0)$, $(2,0)$ and $(2,2)$. A solid lies above $D$ and below the surface $z=4-x^2$.

    1. Write the area of $D$ as a double integral and evaluate it.
    2. Check that $z\ge 0$ on $D$, then write the volume of the solid as a double integral and evaluate it.
11. Q11Mass and Centre of Mass of a Lamina

    A thin plate (a lamina) occupies the triangle $D$ with vertices $(0,0)$, $(1,0)$ and $(1,2)$, where $x$ and $y$ are in centimetres. Its density at $(x,y)$ is $\delta (x,y)=6x+6y$ grams per square centimetre. Find the mass of the plate and the coordinates $(\overline{x},\overline{y})$ of its centre of mass.

12. Q12Synthesis — drawing on several topics in this unit

    A lamina occupies the region $D$ in the first quadrant that lies inside the circle $x^2+y^2=4$ and between the $x$-axis and the line $y=x$. Lengths are in centimetres and the density is $\delta (x,y)=y$ grams per square centimetre.

    1. Describe $D$ in polar coordinates, and also by inequalities of the form $c\le y\le d$, $h_1(y)\le x\le h_2(y)$.
    2. Find the mass of the lamina using polar coordinates.
    3. Set up the mass as a Cartesian iterated integral using your second description, and evaluate it to confirm b).

The 8 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

The rest of this unit

The worksheet is the practice, and it is free. Two more printable documents cover the same unit and come with the University Calculus III bundle: read the notes first, work this sheet, then sit the test closed-book. See what each one covers.

What the Double Integrals notes cover  About the Double Integrals unit test

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

A double integral over a rectangle is two single integrals

Q1 gives a polynomial on a rectangle and asks for the double integral twice: once integrating in y first, once in x first. The inner integral is a partial integration — the other variable is held constant, exactly as it was held constant for a partial derivative — and what it returns is a function of the outer variable alone, which the outer integral then finishes.

Q2 takes that theorem as permission. Both orders give the same value, so the question is only which one is shorter — and it asks you to say which you chose and why. Before integrating anything, try each variable as the inner one: hold the other constant and ask whether the numerator is, up to a constant, the derivative of the bracket in the denominator with respect to the inner variable. In one order it is, and the inner integral is a one-line substitution; in the other it is not.

General regions: the limits come from a sketch

Over a region that is not a rectangle, the inner limits are no longer numbers: they are the curves where a strip enters and leaves the region. Q3 is built on a figure — a parabola and a line enclosing a region, shown in the PDF — and part (a) asks for exactly the two facts the limits need: where the curves meet, and which one is on top in between.

Q4 gives a triangle by its three vertices and asks for the other kind of strip: horizontal, with y on the outside and the left and right boundaries written as x in terms of y. It also asks you to explain in one sentence why that description is the convenient one here. Draw one vertical strip and one horizontal strip across the triangle, and count how many different formulas each needs for its two ends as it sweeps across the region. That count is the sentence.

Reversing the order of integration

Q5 gives an iterated integral in the order dy dx and asks for the same integral in the order dx dy. The limits you are given are a description of a region; the job is to recover the region and describe it again the other way. Part (a) is the sketch, and it is not optional: every reliable reversal goes through one.

Part (c) is the check the whole set leans on: choose a simple integrand, evaluate the integral in both orders, and see that they agree. If they do not, the limits are wrong, not the theorem.

Q6 shows why reversing is worth learning. The integrand has no elementary antiderivative in x, so the integral cannot even be started in the order it is written. Reverse it — same method, sketch first — and the inner integral becomes an integral in y of something that does not contain y at all. Look at what that inner integral multiplies the integrand by: it is exactly the factor a substitution in x needs.

Polar coordinates as a tool

Q7 is the preparation for integrating in polar coordinates: turning Cartesian curves and regions into polar ones. The conversion is always the same three facts — x = r cos θ, y = r sin θ, and x² + y² = r² — substituted into the equation and simplified until r is alone.

Part (b) is a vertical line, which in polar form is not a constant: r depends on θ. Part (c) uses both. The region is part of a disc cut off by that line, and a ray from the origin at angle θ crosses the line first and the circle second. So the inner limits in r run from the line's polar equation to the circle's, and the outer limits in θ are the angles of the two points where the line meets the circle.

Double integrals in polar coordinates

When a region is bounded by circles centred at the origin and by rays from it, polar coordinates turn it into a polar rectangle — constant limits in r and in θ. Q8 is that case: a quarter of a ring. Describe it, substitute for x and y, and replace dA.

Once the limits are constants and the integrand is a function of θ times a function of r, the double integral splits into a product of two single integrals, which is the quickest way to finish Q8.

Q9 hands you a Cartesian iterated integral and asks you to sketch its region, convert and evaluate. The limits describe a line through the origin and a circle; the line through the origin is a ray, θ = constant, and the circle centred at the origin is r = constant. Notice also where the x-range stops and what the left edge of the region is. Then look at the integrand: it depends on x and y only through x² + y², which is r², and together with the r from dA the inner integral becomes a one-line substitution.

Area and volume

Q10 uses one triangle twice. The area of a region is the double integral of 1 over it, and a triangle whose area you already know from geometry is a free check on your limits. The volume of the solid above the region and below a surface z = f(x, y) is the double integral of the height f, provided the surface is on or above the xy-plane throughout the region.

Mass and centre of mass of a lamina

A thin plate whose density varies from point to point is a double integral waiting to happen. Q11 gives a triangular plate, in centimetres, with a density δ(x, y) in grams per square centimetre that grows across it, and asks for its mass and the point where it balances.

The synthesis question

Q12 describes one lamina — the part of a disc in the first quadrant between the x-axis and a line through the origin, with density δ = y — and asks for its mass twice. Part (a) is Q7 and Q4 together: a polar description, which is a polar rectangle because the line through the origin is a ray, and a Cartesian one with horizontal strips, where each strip starts on the line and ends on the circle. Part (b) is Q8's method: substitute, include the r from dA, and split the integral as a product. Part (c) sets up the same mass in Cartesian form from your horizontal-strip description and evaluates it; the inner integral is in x of something that does not contain x, and the outer one needs a substitution. The two answers must agree — which is the whole point of doing it twice.

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Getting the most out of it

Sketch the region before you write a limit

Every question on this sheet that went wrong in someone's hands went wrong in the limits. A thirty-second sketch with the boundary curves labelled and the corner points marked settles which curve is on top, where a strip enters and leaves, and whether the region needs splitting.

Read the limits back as inequalities

Before integrating, write the region you think you have as a pair of inequalities and test one point you know is inside and one you know is outside. If the inequalities disagree with the sketch, the integral will too.

Check the order you chose with a second order

When you reverse the order or switch to polar coordinates, evaluate both versions with an easy integrand such as 1 or x. Equal answers mean the two descriptions are the same region; unequal answers find a wrong limit before it costs you a whole question.

Write dA out in full every time

Write dA = dy dx, dx dy or r dr dθ explicitly in each integral rather than leaving it implicit. It is the fastest way to remember the r in polar coordinates, and to keep the inner limits attached to the right variable.

Estimate before you trust

The integrand lies between its smallest and largest values on the region, so the integral lies between those values times the area. A one-line bound catches a missing factor, a wrong sign or a limit substituted for the wrong variable.

Common questions