Electricity-And-Magnetism · Unit 1 · Video 5 · Interactive Practice

Double Integrals: Area, Charge and Volume

IKey Formulas

Formula	Name	What each element carries
$A = \displaystyle\int_{y_{\min}}^{y_{\max}} \left[\int_{x_1(y)}^{x_2(y)} dx\right] dy$	Area as an iterated integral	Its own area, $dA = dx\,dy$
$Q = \displaystyle\iint \sigma(x,y)\,dx\,dy$	Total charge on a surface	Charge $dQ = \sigma\,dx\,dy$
$dA_1 = r\,d\theta\,dr, \quad dA_2 = 2\pi r\,dr$	Polar patch and thin ring	Area of a patch, or of a ring when $\sigma = \sigma(r)$
$V = \displaystyle\iint f(x,y)\,dx\,dy$	Volume under a surface	Volume of a column of height $f$

Key Insight: The inner integral holds the outer variable fixed and adds one row (or one ring); the outer integral adds the rows. The integrand is whatever one small element carries.

IIArea, One Row at a Time

The inner integral gives the length $x_2(y) - x_1(y)$ of one row; the outer integral stacks the rows into the area.

Region Row height y 1.20

IIICharge as a Limit of Sums

For $\sigma = kxy$ with $k = 1\ \text{C/m}^4$ on the $3 \times 2$ rectangle, the sums $\sum \sigma\,\Delta A$ close in on $Q = \tfrac{k a^2 b^2}{4} = 9\ \text{C}$ as the cells shrink.

Cells per side N 4 Sample σ at

💡 Because $\sigma = kxy$ grows toward the far corner, lower-left samples always undercount and upper-right samples always overcount; the integral is the one value both sums approach.

IVWhy the Polar Patch Carries a Factor $r$

A patch spanning the same angle $d\theta$ is longer far from the centre: its arc is $r\,d\theta$ , so $dA_1 = r\,d\theta\,dr$ .

Radius r 1.20 Element

💡 The ring shortcut $dQ = \sigma(r)\,2\pi r\,dr$ is valid only when every point of the ring has the same density; for $\sigma(r,\theta)$ the $\theta$ integral must be done with $\sigma$ inside it.

VQuiz Questions

Problem 1 · Charge on a Rectangle

Given: A rectangular plate with corners at $(0,0)$ and $(a, b) = (2\ \text{m}, 3\ \text{m})$ carries surface charge density $\sigma = kxy$ with $k = 4\ \text{C/m}^4$ . Find the total charge $Q$ .

✅ Correct!

Q = \tfrac{k a^2 b^2}{4} = \tfrac{4 \cdot 4 \cdot 9}{4} = 36\ \text{C}

❌ Not quite.

\int_0^a x\,dx = \tfrac{a^2}{2}

, not

a^2

: each of the two integrations contributes a factor

\tfrac12

❌ Not quite. The integrals give

Q = 9k

; you still have to multiply by

k = 4\ \text{C/m}^4

, so

Q = 36\ \text{C}

❌ Not quite.

k\,ab

treats

k

as if it were the density. The density is

kxy

, which varies over the plate and must be integrated.

❌ Not quite. Do the inner

x

integral with

y

held fixed, then the outer

y

integral; there are only two factors of

\tfrac12

Show solution

Inner integral, $y$ fixed along the row:

\int_0^2 kxy\,dx = ky\,\frac{2^2}{2} = 2ky

Outer integral adds the rows:

Q = \int_0^3 2ky\,dy = 2k\,\frac{3^2}{2} = 9k = 9 \cdot 4 = 36\ \text{C}

This agrees with the general result $Q = \tfrac{k a^2 b^2}{4} = \tfrac{4 \cdot 2^2 \cdot 3^2}{4} = 36\ \text{C}$ .

Problem 2 · The Missing Factor r

Given: A disc of radius $R = 2\ \text{m}$ carries $\sigma = kr^2$ with $k = 3\ \text{C/m}^4$ . Find the total charge $Q$ .

✅ Correct!

Q = \tfrac{\pi k R^4}{2} = \tfrac{\pi \cdot 3 \cdot 16}{2} = 24\pi \approx 75.4\ \text{C}

❌ Not quite. That is

\iint kr^2\,d\theta\,dr

, which omits the factor

r

dA_1 = r\,d\theta\,dr

. Without it,

d\theta\,dr

is only a length.

❌ Not quite. That is

\sigma(R) \cdot \pi R^2

, which uses the rim density everywhere. The density is smaller nearer the centre.

❌ Not quite. The

\theta

integral runs from

0

2\pi

and contributes a factor

2\pi

, not

1

Show solution

Each element carries $dQ = \sigma\,dA_1 = kr^2 \cdot r\,d\theta\,dr$ . Do the $\theta$ integral first ( $r$ fixed):

Q = \int_0^R \left[\int_0^{2\pi} kr^3\,d\theta\right] dr = \int_0^R 2\pi k r^3\,dr = \frac{2\pi k R^4}{4} = \frac{\pi k R^4}{2}

Q = \frac{\pi \cdot 3 \cdot 2^4}{2} = 24\pi \approx 75.4\ \text{C}

Problem 3 · Rows with Moving Ends

Given: A region in the $xy$ plane: the row at height $y$ runs from $x_1(y) = y^2$ to $x_2(y) = 2y$ , for $0 \le y \le 2$ (lengths in m). Find the area $A$ of the region.

✅ Correct! The row length is

2y - y^2

, and

\int_0^2 (2y - y^2)\,dy = \tfrac43\ \text{m}^2

❌ Not quite.

\int_0^2 2y\,dy = 4

measures each row from

x = 0

. The row starts at

x_1(y) = y^2

, so subtract it.

❌ Not quite. The inner integral gives

x_2(y) - x_1(y)

, a difference, not a sum.

❌ Not quite.

\int_0^2 y^2\,dy = \tfrac83

is the area to the left of the region. The row length is

x_2(y) - x_1(y)

Show solution

Inner integral: the length of the row at height $y$ ,

\int_{y^2}^{2y} dx = 2y - y^2

Outer integral adds the rows from $y_{\min} = 0$ to $y_{\max} = 2$ :

A = \int_0^2 (2y - y^2)\,dy = \left[y^2 - \frac{y^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}\ \text{m}^2

Check: $x_2(y) \ge x_1(y)$ on $[0, 2]$ since $2y - y^2 = y(2 - y) \ge 0$ , and the two edges meet at $y = 0$ and $y = 2$ .

Problem 4 · When the Ring Shortcut Fails

Given: A disc of radius $R = 1\ \text{m}$ carries $\sigma(r, \theta) = k\,r\,(1 + \cos\theta)$ with $k = 3\ \text{C/m}^3$ , so it is denser on the right ( $\theta = 0$ ) than on the left. Find the total charge $Q$ .

✅ Correct! The

\cos\theta

part integrates to zero around each ring, leaving

Q = \tfrac{2\pi k R^3}{3} = 2\pi\ \text{C}

❌ Not quite. That uses the density at

\theta = 0

\sigma = 2kr

, for the whole ring. Points on one ring carry different densities here, so the ring shortcut does not apply.

❌ Not quite. That omits the factor

r

in the area element

r\,d\theta\,dr

❌ Not quite. Only the

\cos\theta

term averages to zero around a ring; the constant

1

1 + \cos\theta

does not, and

\sigma \ge 0

everywhere.

Show solution

With no symmetry in $\theta$ , do the full integral with $\sigma(r,\theta)$ inside the $\theta$ integral:

Q = \int_0^R \left[\int_0^{2\pi} k\,r\,(1 + \cos\theta)\,r\,d\theta\right] dr

At fixed $r$ : $\int_0^{2\pi} (1 + \cos\theta)\,d\theta = 2\pi + 0 = 2\pi$ , so the bracket is $2\pi k r^2$ .

Q = \int_0^R 2\pi k r^2\,dr = \frac{2\pi k R^3}{3} = \frac{2\pi \cdot 3 \cdot 1}{3} = 2\pi \approx 6.28\ \text{C}

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Double Integrals: Area, Charge and Volume

IKey Formulas

IIArea, One Row at a Time

IIICharge as a Limit of Sums

IVWhy the Polar Patch Carries a Factor rrr

VQuiz Questions

IVWhy the Polar Patch Carries a Factor $r$