Unit 2

Applications of Integration

Every application in this unit is the same three-step move: slice the thing into pieces so thin that each one is simple, write down the contribution of a single slice, then integrate. If you can write one slice correctly, the calculus is bookkeeping. Almost every lost mark in this unit is a bad slice, not a bad integral.

1 · The slicing principle

Do not memorise ten formulas. Memorise one procedure and rebuild whichever formula you need — that way you can also handle the problem that does not match any formula you were taught.

Draw the region. Not optional. Nearly every error in this unit is visible in a correct sketch.
Slice it into strips of thickness \(dx\) (or \(dy\), or \(dt\)).
Write the contribution of one slice — its area, volume, work, or charge — entirely in terms of the slicing variable.
Integrate over the range that sweeps the slice across the whole region.

Choosing \(dx\) or \(dy\)

Slice perpendicular to the axis whose variable you are integrating. Pick whichever direction lets a single formula describe every slice. If a horizontal strip changes which curve bounds it partway up the region, slice vertically instead — or split the integral. Choosing well often turns two integrals into one.

2 · Area between curves

\[ A = \int_a^b \big[f(x) - g(x)\big]\,dx \] Top minus bottom, always. If the curves cross inside \([a,b]\), split at the crossing points and use top − bottom separately on each piece.

For regions that are easier to describe sideways, integrate in \(y\) with right minus left:

\[ A = \int_c^d \big[x_R(y) - x_L(y)\big]\,dy \]

Never integrate through a crossing point

If you integrate \(f-g\) straight through a place where the curves swap order, the piece where \(g\) is on top contributes a negative area and partially cancels the rest. You get a number, it just is not the area. Find the intersections first, then handle each subinterval with the correct ordering.

3 · Volumes of revolution

Disks and washers — slice ⟂ to the axis of rotation

\[ V = \pi\int_a^b \big[R(x)\big]^2 dx \qquad\qquad V = \pi\int_a^b \Big(\big[R_{\text{out}}\big]^2 - \big[R_{\text{in}}\big]^2\Big)\,dx \] Each slice is a coin (disk) or a coin with a hole (washer). \(R\) is measured from the axis of rotation, not from the \(x\)-axis — which matters the moment the axis is \(y = 2\) rather than \(y = 0\).

Squares do not subtract

The washer integrand is \(R_{\text{out}}^2 - R_{\text{in}}^2\), not \(\big(R_{\text{out}} - R_{\text{in}}\big)^2\). Those are different numbers. This single mistake costs more exam marks than anything else in the unit.

Shells — slice ∥ to the axis of rotation

\[ V = 2\pi\int_a^b (\text{radius})(\text{height})\,dx \;=\; 2\pi\int_a^b x\,f(x)\,dx \] Each slice is a thin cylindrical can. Unroll it and you get a rectangle: circumference \(2\pi r\), times height, times thickness.

Same region, same strip, different axis. Perpendicular to the axis ⇒ disk/washer. Parallel to the axis ⇒ shell. That one sentence decides the method every time.

Which method?

Both always work; one is usually far easier. Rotating a region bounded by \(y = f(x)\) about the \(y\)-axis is natural for shells — no need to solve for \(x\). About the \(x\)-axis it is natural for disks. If a problem is forcing you to invert an awkward function, you probably chose the harder method; switch.

4 · Arc length and surface area

\[ L = \int_a^b \sqrt{1 + \big[f'(x)\big]^2}\;dx \] Straight from Pythagoras on an infinitesimal piece: \(dL = \sqrt{dx^2 + dy^2} = \sqrt{1 + (dy/dx)^2}\,dx\).

\[ S = 2\pi\int_a^b f(x)\sqrt{1 + \big[f'(x)\big]^2}\;dx \] Surface of revolution about the \(x\)-axis: circumference \(2\pi y\) times the arc length element. Note it uses \(dL\), not \(dx\) — that radical is the difference between a surface area and a wrong answer.

Be warned: arc-length integrands are radicals of polynomials, and most have no elementary antiderivative. Textbook problems are rigged so that the expression under the root becomes a perfect square. If yours does not, either you differentiated wrong or the problem wants a numerical answer.

5 · Work

\[ W = \int_a^b F(x)\,dx \] Force times distance, when the force varies. Constant force is the special case that needs no integral.

The last row is the same integral as the first. A capacitor is a spring for charge.
Situation	Slice reasoning	Result
Stretching a spring	Hooke's law \(F = kx\)	\(W = \displaystyle\int_0^d kx\,dx = \tfrac12 kd^2\)
Pumping liquid from a tank	A slice of thickness \(dy\) at height \(y\) weighs \(\rho g A(y)\,dy\) and is lifted \((h-y)\)	\(W = \displaystyle\int \rho g A(y)(h-y)\,dy\)
Lifting a hanging cable	Each piece travels a different distance	\(W = \displaystyle\int (\text{density})(g)(\text{distance})\,dy\)
Charging a capacitor	Moving \(dq\) across a voltage that is itself rising, \(v = q/C\)	\(W = \displaystyle\int_0^Q \frac{q}{C}\,dq = \frac{Q^2}{2C}\)

6 · Average value

\[ f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx \] Total accumulated, divided by the width. Geometrically: the height of the rectangle with the same area as the region under the curve.

The Mean Value Theorem for Integrals says a continuous \(f\) actually attains its average somewhere: there is a \(c\) in \([a,b]\) with \(f(c) = f_{\text{avg}}\).

7 · RMS and average power

Here is where Calculus II stops being coursework and becomes your job. The average value of a sine wave over a full cycle is zero — yet a wall socket plainly delivers power. The average is simply the wrong statistic.

Power in a resistor is \(p = v^2/R\), which depends on \(v\)². So average the square, then take the root to get back to volts:

\[ V_{\text{rms}} = \sqrt{\frac{1}{T}\int_0^{T} \big[v(t)\big]^2 dt} \] Read the name backwards: Root of the Mean of the Square. That is literally the formula, right to left.

RMS is defined so that a DC voltage of \(V_{\text{rms}}\) would heat a resistor at exactly the same rate as the AC waveform does. It is an equivalent-heating voltage — which is why every meter and every spec sheet quotes it.

Squaring folds the negative half up. The squared wave has a genuine nonzero average — half the peak — and the square root of that is the RMS value. Note the squared wave oscillates at twice the original frequency, which is why AC power delivery pulses at 120 Hz on a 60 Hz line.

The square wave has the highest RMS for a given peak — it spends all its time at full amplitude. Every entry here is one integral you can do yourself.
Waveform, peak \(V_p\)	RMS value	Average value
Sine	\(V_p/\sqrt{2} \approx 0.707\,V_p\)	\(0\)
Square, \(\pm V_p\)	\(V_p\)	\(0\)
Triangle, \(\pm V_p\)	\(V_p/\sqrt{3} \approx 0.577\,V_p\)	\(0\)
Sawtooth, \(0\) to \(V_p\)	\(V_p/\sqrt{3}\)	\(V_p/2\)
Half-wave rectified sine	\(V_p/2\)	\(V_p/\pi\)
Full-wave rectified sine	\(V_p/\sqrt{2}\)	\(2V_p/\pi\)
DC level \(V_0\)	\(V_0\)	\(V_0\)

\[ P_{\text{avg}} = \frac{V_{\text{rms}}^2}{R} = I_{\text{rms}}^2 R = V_{\text{rms}} I_{\text{rms}}\cos\phi \] The \(\cos\phi\) is the power factor, present when voltage and current are out of phase. For a pure resistor \(\phi = 0\) and it drops out.

Why "120 V" is really 170 V

A North American wall outlet is quoted as 120 V RMS. Its actual peak is \(120\sqrt{2} \approx 170\) V, and it swings from \(+170\) V to \(-170\) V. Insulation must be rated for the peak; heating is governed by the RMS. Confusing the two is a design error with consequences, not just a lost mark.

8 · Charge and stored energy

The three fundamental circuit relations are integrals. They come straight from the slicing principle: accumulate a rate over time.

\[ q(t) = \int_0^t i(\tau)\,d\tau + q(0) \] \[ v_C(t) = \frac{1}{C}\int_0^t i\,d\tau + v_C(0) \qquad i_L(t) = \frac{1}{L}\int_0^t v\,d\tau + i_L(0) \] The constant of integration is not a formality here — it is the initial condition, the charge or current already present when you started watching. Dropping it is dropping physics.

Deriving the energy formulas

You have probably been handed \(\tfrac12 CV^2\) and \(\tfrac12 LI^2\) to memorise. They are one-line integrals. Energy is accumulated power, and power is \(vi\):

\[ W = \int p\,dt = \int v\,i\,dt \]

Capacitor. Substitute \(i = C\,dv/dt\). The \(dt\) cancels and the variable of integration becomes \(v\):

\[ W = \int v\cdot C\frac{dv}{dt}\,dt = C\int_0^V v\,dv = \tfrac12 CV^2 \]

Inductor. Identical structure with \(v = L\,di/dt\):

\[ W = \int L\frac{di}{dt}\cdot i\,dt = L\int_0^I i\,di = \tfrac12 LI^2 \]

Both are \(\tfrac12(\cdots)^2\) for the same reason a spring stores \(\tfrac12 kx^2\): you are integrating something that grows linearly from zero, and the average of a ramp is half its final value.

9 · Centroids and moments

For a thin plate of uniform density bounded above by \(f\) and below by \(g\), with area \(A\):

\[ \bar{x} = \frac{1}{A}\int_a^b x\big[f-g\big]\,dx \qquad\qquad \bar{y} = \frac{1}{A}\int_a^b \tfrac12\big[f^2 - g^2\big]\,dx \] Each is a weighted average of position. The \(\tfrac12[f^2-g^2]\) in \(\bar y\) is the strip's own midpoint times its height.

Symmetry shortcut: if the region is symmetric about a line, the centroid lies on that line. Use it before integrating — it frequently kills one of the two integrals outright.

The same weighted-average machinery, with the weight \(x\) replaced by \(x^2\), gives the moment of inertia — and with a probability density in place of the geometry, gives expected value. Once you see the pattern \(\int(\text{quantity})(\text{weight})\), it recurs everywhere.

10 · Worked examples

1. Washer volume

The region bounded by \(y = x^2\) and \(y = 2x\) is rotated about the \(x\)-axis. Find the volume.

Intersections: \(x^2 = 2x\) gives \(x = 0\) and \(x = 2\).

On \((0,2)\), \(2x > x^2\) — check at \(x=1\): \(2 > 1\) ✓ — so the line is the outer radius and the parabola the inner.

\[ V = \pi\int_0^2 \Big[(2x)^2 - (x^2)^2\Big]dx = \pi\int_0^2 \big(4x^2 - x^4\big)dx \] \[ = \pi\left[\frac{4x^3}{3} - \frac{x^5}{5}\right]_0^2 = \pi\left(\frac{32}{3} - \frac{32}{5}\right) = \frac{64\pi}{15} \]

Sanity check: the answer must be positive and smaller than the solid swept by the line alone, \(\pi\int 4x^2 = 32\pi/3 \approx 33.5\). We got \(\approx 13.4\). ✓

2. Shell volume

The region under \(y = e^{-x^2}\) from \(x=0\) to \(1\) is rotated about the \(y\)-axis. Find the volume.

Disks would need \(x\) as a function of \(y\) — solving \(y = e^{-x^2}\) for \(x\) is ugly. Shells avoid that entirely.

\[ V = 2\pi\int_0^1 x\,e^{-x^2}\,dx \]

Substitute \(u = -x^2\), \(du = -2x\,dx\), so \(x\,dx = -du/2\):

\[ = 2\pi\cdot\left(-\tfrac12\right)\int_0^{-1} e^{u}\,du = -\pi\Big[e^{u}\Big]_0^{-1} = \pi\left(1 - \tfrac1e\right) \approx 1.986 \]

Note the payoff: the \(x\) supplied by the shell formula is exactly the factor that makes the substitution work. \(\int e^{-x^2}dx\) alone has no elementary antiderivative — but \(\int x e^{-x^2}dx\) is trivial.

3. RMS of a sine wave, from the definition

Show that \(v(t) = V_p\sin(\omega t)\) has \(V_{\text{rms}} = V_p/\sqrt2\). Take \(T = 2\pi/\omega\), one full period.

\[ V_{\text{rms}}^2 = \frac{1}{T}\int_0^{T} V_p^2\sin^2(\omega t)\,dt \]

Use the half-angle identity — the same one from Unit 1:

\[ \sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2} \] \[ V_{\text{rms}}^2 = \frac{V_p^2}{T}\int_0^{T}\frac{1-\cos(2\omega t)}{2}\,dt = \frac{V_p^2}{2T}\left[t - \frac{\sin(2\omega t)}{2\omega}\right]_0^{T} \]

The sine term vanishes at both ends — \(2\omega T = 4\pi\) is a whole number of cycles. This is the key step: the cosine averages to zero over a full period, leaving only the constant \(\tfrac12\).

\[ V_{\text{rms}}^2 = \frac{V_p^2}{2T}\cdot T = \frac{V_p^2}{2} \quad\Longrightarrow\quad V_{\text{rms}} = \frac{V_p}{\sqrt2} \]

Note what did not matter: \(\omega\) cancelled. The RMS of a sinusoid is independent of frequency — 60 Hz or 60 MHz, always peak over \(\sqrt2\).

4. RMS of a half-wave rectified sine

A diode passes only the positive half of \(V_p\sin(\omega t)\); the negative half is clipped to zero. Find the RMS value.

Set up carefully. The period is still \(T\), but the integrand is nonzero only on the first half. Average over the whole period:

\[ V_{\text{rms}}^2 = \frac{1}{T}\left[\int_0^{T/2} V_p^2\sin^2(\omega t)\,dt + \int_{T/2}^{T} 0\,dt\right] \]

From the previous example \(\sin^2\) averages to \(\tfrac12\) over any whole number of half-cycles, so the first integral is \(V_p^2 (T/2)(\tfrac12) = V_p^2 T/4\):

\[ V_{\text{rms}}^2 = \frac{1}{T}\cdot\frac{V_p^2 T}{4} = \frac{V_p^2}{4} \quad\Longrightarrow\quad V_{\text{rms}} = \frac{V_p}{2} \]

Sensible? Half-wave rectification throws away half the energy, so \(V_{\text{rms}}^2\) should halve — and \(V_p^2/4\) is indeed half of \(V_p^2/2\). The RMS voltage drops only by \(1/\sqrt2\) (from \(0.707V_p\) to \(0.5V_p\)) because power goes as the square. Never average the RMS values of two waveforms — average their squares.

5. Charge and energy from a current waveform

A 100 µF capacitor, initially uncharged, is fed \(i(t) = 0.02t\) A for \(0 \le t \le 3\) s. Find \(q(3)\), \(v(3)\), and the energy stored.

\[ q(3) = \int_0^3 0.02t\,dt = 0.02\left[\frac{t^2}{2}\right]_0^3 = 0.02(4.5) = 0.09\ \text{C} \] \[ v(3) = \frac{q}{C} = \frac{0.09}{100\times10^{-6}} = 900\ \text{V} \]

Energy, two ways — both must agree:

\[ W = \tfrac12 CV^2 = \tfrac12(100\times10^{-6})(900)^2 = 40.5\ \text{J} \] \[ W = \frac{Q^2}{2C} = \frac{(0.09)^2}{2(100\times10^{-6})} = \frac{0.0081}{0.0002} = 40.5\ \text{J}\;✓ \]

Watch the units. 100 µF is \(10^{-4}\) F. Leaving capacitance in microfarads is the most common way to be off by a factor of a million here.

6. Arc length

Find the length of \(y = \tfrac23 x^{3/2}\) from \(x = 0\) to \(x = 3\).

\(y' = x^{1/2}\), so \((y')^2 = x\):

\[ L = \int_0^3 \sqrt{1+x}\;dx \]

The radical collapsed to something integrable — the tell that this problem was constructed to work out. Substitute \(u = 1+x\):

\[ = \int_1^4 u^{1/2}\,du = \tfrac23\Big[u^{3/2}\Big]_1^4 = \tfrac23(8-1) = \frac{14}{3} \approx 4.67 \]

Sanity check: the straight-line distance from \((0,0)\) to \((3, 2\sqrt3 \approx 3.46)\) is \(\sqrt{9+12} \approx 4.58\). An arc must be slightly longer than its chord, and \(4.67 > 4.58\). ✓

Checklist

I draw the region before writing any integral.
I use \(R_{\text{out}}^2 - R_{\text{in}}^2\), never \((R_{\text{out}} - R_{\text{in}})^2\).
I choose disks vs shells by whether the strip is perpendicular or parallel to the axis.
I measure radii from the axis of rotation, not from a coordinate axis.
I can derive \(V_{\text{rms}} = V_p/\sqrt2\) from the definition, not just quote it.
I know average value and RMS are different statistics, and why AC needs the second one.
I can derive \(\tfrac12 CV^2\) and \(\tfrac12 LI^2\) by integrating \(vi\).
I keep the constant of integration when it is an initial condition.