Applications of Integration and Polar Curves

Updated 7 October 2026

A definite integral adds up many small pieces, and the choice of piece decides what it measures: thin rectangles give an area, thin discs give a volume, short straight segments give the length of a curve, and thin sectors give the area of a region described by an angle. A differential equation contains a derivative, and solving it means integrating back from the rate of change to the quantity itself.

What does integration measure in practice?

An integral adds up a rate over an interval, so it measures the total change of whatever the rate describes: velocity gives displacement, and acceleration gives change in velocity. A car with v = 10 − 2t meters per second has displacement $10t-t^2$10t − t², so from t = 0 to 5 it moves 50 − 25 = 25 meters. In a short time dt the car moves about v dt, and the integral adds those moves exactly. Negative velocity subtracts, so the integral is net change of position, not distance. For distance, solve v = 0 to find where the direction changes, integrate each part, and add the sizes. From t = 0 to 10 the car's displacement is 100 − 100 = 0, but it travels 25 + 25 = 50 meters. The velocity graph from 0 to 5 is a triangle, and its area, $\frac{1}{2}\times 5\times 10=25$1/2 × 5 × 10 = 25, agrees.

Distance counts every stretch as positive, so it is never less than the size of the displacement.

See Velocity and displacement by integration and Total distance traveled.

How do you find the area between two curves?

Integrate the upper curve minus the lower one between their crossings: area $=\int \left(\text{upper}-\text{lower}\right)dx$= ∫ (upper − lower) dx. For y = x and $y=x^2$y = x², setting $x^2=x$x² = x gives crossings at 0 and 1, with x on top between them. The area is the integral of $x-x^2$x − x² from 0 to 1, which is $\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$1/2 − 1/3 = 1/6. Each strip has height upper − lower and width dx, so neither curve needs to lie above the axis. If the curves swap over, split at every crossing and use the positive difference on each part; otherwise the pieces cancel. For $y=x^3$y = x³ and y = x from −1 to 1, each lobe has area $\frac{1}{4}$1/4, so the total is $\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$1/4 + 1/4 = 1/2, though one integral over the whole range gives 0. As a sanity bound, the region between x and $x^2$x² sits inside the triangle under y = x, whose area $\frac{1}{2}$1/2 exceeds $\frac{1}{6}$1/6.

Sketch the curves first to see where they swap over.

See Area between two curves.

How do you find the volume of a solid of revolution?

Spin the region under a curve about the x-axis and integrate $\pi y^{2}dx$πy² dx, since each thin slice is a disc of radius y and thickness dx. For $y=\sqrt{x}$y = √x from 0 to 4, $y^2$y² is x, and the integral of x is $\frac{16}{2}=8$16/2 = 8, so the volume is $8\pi$8π. Every slice is a cylinder whose volume is its face area times its thickness, and the integral adds the slices. About the y-axis the radius is x and the thickness dy, so integrate $\pi x^{2}dy$πx² dy with limits in y. A region that does not reach the axis makes rings, with area $\pi \left(R^2-r^2\right)$π(R² − r²); square each radius before subtracting. A cone makes a good test. Spinning y = x from 0 to 3 gives $\pi$π times the integral of $x^2$x², and $\frac{27}{3}=9$27/3 = 9, so the volume is $9\pi$9π, one third of the $27\pi$27π cylinder of the same size.

So the washer volume is $V=\pi \int \left(R^2-r^2\right)dx$V = π ∫ (R² − r²) dx.

A solid need not be made by spinning. If every slice has a known shape, integrate the slice area: $V=\int A\left(x\right)dx$V = ∫ A(x) dx. Squares standing on the region under $y=\sqrt{x}$y = √x have area x, so on [0, 4] the volume is $\int xdx=8$∫ x dx = 8, and no $\pi$π appears.

See Volumes of revolution, Volumes by washers and Volumes with known cross sections.

How do you find the length of a curve, or its average height?

For the length of a curve, integrate $\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}$√(1 + (dy/dx)²) in x. A small piece of curve is the hypotenuse of a right triangle with legs dx and dy, so its length is $\sqrt{d{x}^2+d{y}^2}$√(dx² + dy²), and taking dx outside the root gives the formula. For $y=\frac{4x}{3}$y = 4x/3 from 0 to 3, the root is $\sqrt{\frac{25}{9}}$√(25/9), since $\frac{9}{9}+\frac{16}{9}=\frac{25}{9}$9/9 + 16/9 = 25/9, and the length is $\frac{5}{3}\times 3=5$5/3 × 3 = 5, the hypotenuse of a 3, 4, 5 triangle. Most curves give a root with no antiderivative, and then the length is found numerically. The average height on an interval is the area divided by the width. On $x^2$x² from 0 to 3, the area is $\frac{27}{3}=9$27/3 = 9, so the average is 9 ÷ 3 = 3: a rectangle that tall on the same base holds the same area. That average sits between the lowest value, 0, and the highest, 9.

This average height is called the mean value of the function.

See Arc length and The mean value of a function.

What if the region never ends?

An improper integral has an infinite limit, or an integrand that becomes infinite at one end. Replace the troublesome limit with a variable t, integrate to t, then let t go to infinity. Under $\frac{1}{x^2}$1/x² from 1 to t the area is $1-\frac{1}{t}$1 − 1/t, which tends to 1, so a region of endless width has finite area; at t = 4 it is $1-\frac{1}{4}=\frac{3}{4}$1 − 1/4 = 3/4. The limit is the only meaning an endless sum of strips can have. Not every such area is finite: under $\frac{1}{x}$1/x the area is ln t, which grows without bound, so that integral diverges, though $\frac{1}{x}$1/x also shrinks to 0. An infinite integrand is treated the same way at its end, so the area under $\frac{1}{\sqrt{x}}$1/√x from t to 1 is $2-2\sqrt{t}$2 − 2√t, which tends to 2. The pieces of the first area agree: $\frac{1}{2}$1/2 from 1 to 2, since $1-\frac{1}{2}=\frac{1}{2}$1 − 1/2 = 1/2, and $\frac{1}{2}$1/2 beyond 2.

The same distinction between converging and diverging runs through sequences and series.

See Improper integrals.

What if the integral cannot be done at all?

Some functions, such as $e^{-x^2}$e^(−x²), have no antiderivative in standard functions, yet the area under them is a definite number, so estimate it numerically. The trapezium rule gives each strip of width h a slanted top: $\frac{h}{2}$h/2 times the first height, plus twice each inner height, plus the last. For $x^2$x² from 0 to 2 with h = 1, the heights are 0, 1 and 4, so it gives $\frac{1}{2}\times \left(0+2+4\right)=3$1/2 × (0 + 2 + 4) = 3, against the exact $\frac{8}{3}$8/3. Chords cut across the bend, so on this upward curve the rule overestimates, and halving h cuts the error to about a quarter. Simpson's rule fits a parabola to each pair of strips, so it needs an even number, with weights 1, 4, 1 times $\frac{h}{3}$h/3 for one pair. It gives $\frac{1}{3}\times \left(0+4+4\right)=\frac{8}{3}$1/3 × (0 + 4 + 4) = 8/3, exact for a parabola. The trapezium error is $3-\frac{8}{3}=\frac{1}{3}$3 − 8/3 = 1/3, and with $h=\frac{1}{2}$h = 1/2 it falls to $\frac{1}{12}$1/12.

In symbols the trapezium rule is $\left(\frac{h}{2}\right)\left(y_0+2y_1+\cdots +2y_{n-1}+y_n\right)$(h/2)(y₀ + 2y₁ + … + 2yₙ₋₁ + yₙ).

With more strips the Simpson weights run 1, 4, 2, 4, …, 4, 1, multiplied by $\frac{h}{3}$h/3, and the rule is exact for any cubic.

See Numerical integration and Simpson's rule.

What is a differential equation?

Any equation containing a derivative is a differential equation, and its solution is a function. Forming one is a translation: "grows at a rate proportional to its size" becomes $\frac{dP}{dt}=$dP/dt = kP. If the variables separate, put every P on one side with dP and every t on the other, then integrate: $\frac{dP}{P}=kdt$dP/P = k dt gives ln P = kt + c, so $P=A{e}^{kt}$P = Ae^(kt). A known value fixes the constant. With P = 100 at t = 0 and k = ln 2, P doubles each unit of time and reaches 100 × 2 × 2 × 2 = 800 at t = 3. Integrating undoes the differentiation that built the equation. An equation like $\frac{dy}{dx}+y=x$dy/dx + y = x does not separate, so multiply by the integrating factor $e^x$eˣ, which makes the left side the single derivative of yeˣ. And $P=100\times 2^t$P = 100 × 2ᵗ has derivative (ln 2) P, which is kP.

A known value of y at a known x is called an initial condition.

A linear equation $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$dy/dx + P(x) y = Q(x) does not separate. Multiply through by the integrating factor $I=e^{\int Pdx}$I = e^(∫ P dx). The left side becomes $\frac{d}{dx}\left(Iy\right)$d/dx (Iy), a single derivative, so integrate both sides and divide by I.

See Forming differential equations, Separable differential equations and The integrating factor.

What if you cannot solve it exactly?

When a differential equation cannot be solved exactly, draw its slope field or step through it numerically. A slope field puts a short segment at each point with the gradient the equation gives there; for $\frac{dy}{dx}=x-y$dy/dx = x − y, the segment at (2, 1) has gradient 2 − 1 = 1. A solution follows the segments wherever it goes, so the field shows every solution at once. Euler's method turns this into numbers: from a known point, follow the gradient for a step h, find the new gradient, and repeat. With y' = y, starting at y = 1 when x = 0 and taking $h=\frac{1}{2}$h = 1/2, the steps give $\frac{3}{2}$3/2 and then $\frac{3}{2}\times \frac{3}{2}=\frac{9}{4}$3/2 × 3/2 = 9/4 at x = 1, below the exact e. An upward-bending curve lies above its tangents, so the estimate lands low; a smaller step shrinks the miss, and a downward bend flips its sign. Since e lies between $\frac{5}{2}$5/2 and 3, $\frac{9}{4}$9/4 is indeed too low.

Each Euler step reads the gradient only at its start, which is where the error comes from.

See Slope fields and Euler's method.

How do you solve a second-order equation?

For ay′′ + by′ + cy = 0, try $y=e^{mx}$y = e^(mx): every derivative is a multiple of $e^{mx}$e^(mx), so dividing it out leaves the auxiliary equation $a{m}^2+bm+c=0$am² + bm + c = 0. For y'' − 5y' + 6y = 0 the auxiliary equation $m^2-5m+6=0$m² − 5m + 6 = 0 has roots 2 and 3, so $y=A{e}^{2x}+$y = Ae^(2x) + Be${}^{3x}$^(3x). An exponential keeps its shape under differentiation, so the equation becomes algebra. Two initial values fix A and B: y = 0 and y' = 1 at x = 0 give A + B = 0 and 2A + 3B = 1, so A = −1 and B = 1. The kind of root changes the form. A repeated root m gives $\left(A+Bx\right)e^{mx}$(A + Bx)e^(mx), and complex roots $p\pm qi$p ± qi give $e^{px}\left(A\cos qx+B\sin qx\right)$e^(px)(A cos qx + B sin qx), an oscillation. Both roots satisfy the auxiliary equation: 4 − 10 + 6 = 0 and 9 − 15 + 6 = 0.

In general, two real roots $m_1$m₁ and $m_2$m₂ give $y=A{e}^{m_{1}x}+B{e}^{m_{2}x}$y = A e^(m₁x) + B e^(m₂x).

With a nonzero right side f(x), the complementary function solves the equation with 0 on the right, and the particular integral produces f(x): try a function of the same form with unknown coefficients, substitute, and match coefficients. Add the two.

See The auxiliary equation and Complementary function and particular integral.

Why does this describe oscillation?

In simple harmonic motion the acceleration is proportional to the displacement and points back to the center, ẍ $=-{\omega }^{2}x$= −ω²x. The auxiliary equation $m^2+{\omega }^2=0$m² + ω² = 0 has roots $\pm \omega i$±ωi, so $x=A\cos \omega t+B\sin \omega t$x = A cos ωt + B sin ωt, which is $R\cos \left(\omega t-\alpha \right)$R cos(ωt − α) with amplitude $R=\sqrt{A^2+B^2}$R = √(A² + B²) and period $\frac{2\pi }{\omega }$2π/ω. With $\omega =2$ω = 2, A = 3 and B = 4, the amplitude is 5, because 9 + 16 = 25, and the period is $\pi$π. The period does not depend on R: a body pulled twice as far is pushed back twice as hard, so it covers twice the distance in the same time. Resistance growing with speed adds a term kẋ, and the discriminant $k^2-4{\omega }^2$k² − 4ω² decides the motion: negative gives a shrinking oscillation, positive a slow return with none. Differentiating x = 3 cos 2t twice gives −12 cos 2t, and −4 × 3 = −12 matches $-{\omega }^{2}x$−ω²x.

Critical damping, where the discriminant is 0, is the fastest return that never crosses the center.

See Simple harmonic motion and Damped oscillations.

What if two quantities depend on each other?

Two quantities are coupled when each one's rate of change depends on both, as with predators and prey, so neither equation can be solved alone. Write the pair as X' = MX and try $X=e^{\lambda t}v$X = e^(λt) v: this works exactly when v is an eigenvector of M with eigenvalue $\lambda$λ. For x' = y and y' = x, M has rows (0, 1) and (1, 0), with eigenvalue 1 for (1, 1) and −1 for (1, −1), so the general solution is $A{e}^t\left(1,1\right)+B{e}^{-t}\left(1,-1\right)$A e^(t) (1, 1) + B e^(−t) (1, −1). Along an eigenvector the matrix only stretches, so each part changes at a single rate. The signs of the eigenvalues predict the long run, and one of each makes a saddle. The eigenvalues multiply to the determinant, 0 × 0 − 1 × 1 = −1, and 1 × (−1) = −1 agrees.

With two real eigenvalues the general solution is $A{e}^{{\lambda }_{1}t}{v}_1+B{e}^{{\lambda }_{2}t}{v}_2$A e^(λ₁t) v₁ + B e^(λ₂t) v₂, and the initial values fix A and B. The eigenvalue work is in matrices and Markov chains.

A phase portrait plots y against x, so each solution is one path, and the signs of the eigenvalues predict its shape: both negative, every path runs into the origin (a stable node); both positive, every path runs away (an unstable node); complex, the paths spiral.

A second-order equation is a coupled system in disguise: name the derivative $v=\frac{dx}{dt}$v = dx/dt, and one equation becomes the pair $\frac{dx}{dt}=v$dx/dt = v and $\frac{dv}{dt}$dv/dt in terms of x and v. The characteristic polynomial of that matrix is the auxiliary equation.

See Coupled differential equations, Solving a coupled system with eigenvalues, Phase portraits and equilibrium types and A second-order equation as a coupled system.

What are polar coordinates?

Polar coordinates name a point by its distance r from the origin and the angle $\theta$θ from the positive x-axis. The right triangle converts them: $x=r\cos \theta$x = r cos θ and $y=r\sin \theta$y = r sin θ, and back again, $r=\sqrt{x^2+y^2}$r = √(x² + y²) and $\tan \theta =\frac{y}{x}$tan θ = y/x. The point (−3, 4) has r = 5, since 9 + 16 = 25, and lies in the second quadrant, so $\theta$θ is $\pi$π minus the acute angle with tangent $\frac{4}{3}$4/3. It matters because $\tan \theta$tan θ repeats every half turn, so an inverse tangent returns only angles between $-\frac{\pi }{2}$−π/2 and $\frac{\pi }{2}$π/2. A polar curve lets r depend on $\theta$θ: r = 2 is a circle, and $r=2\cos \theta a$r = 2 cos θ a circle through the origin. A region swept by an angle is built from sectors, so its area is $\frac{1}{2}\int r^{2}d\theta$½ ∫ r² dθ, not $\int ydx$∫ y dx. Converting back gives x as $5\times \left(-\frac{3}{5}\right)=-3$5 × (−3/5) = −3 and y as $5\times \frac{4}{5}=4$5 × 4/5 = 4.

To sketch a polar curve, read r at the quarter turns, mark those points, and find where r reaches 0. Differentiating r with respect to $\theta$θ shows where the curve moves in toward the pole.

A sector of angle $d\theta$dθ is the fraction $\frac{d\theta }{2\pi }$dθ/2π of a disc of radius r, so its area is ½$r^{2}d\theta$r² dθ, and adding the sectors gives area = ½ $\int r^{2}d\theta$∫ r² dθ between the two angles. Check it on a quarter disc of radius 2: ½ $\times 4\times \frac{\pi }{2}=\pi$× 4 × π/2 = π, a quarter of $4\pi$4π. Between two curves, subtract the squares: ½ $\int \left(R^2-r^2\right)d\theta$∫ (R² − r²) dθ.

See Polar coordinates, Sketching polar curves, The rate of change of a polar function and The area of a polar region.

The mistakes worth naming

• Giving displacement when distance was asked. Find where v = 0, integrate between those times, and add the absolute values.
• Subtracting the radii before squaring. The area of a ring is $\pi \left(R^2-r^2\right)$π(R² − r²), not $\pi {\left(R-r\right)}^2$π(R − r)².
• Wrong limits for an enclosed area. Solve for the crossing points first, and split where the curves swap over.
• An odd number of strips in Simpson's rule. One parabola covers each pair of strips, so the strips must pair up.
• Using $\int ydx$∫ y dx for a polar area. A polar region is swept in sectors, so the area is ½ $\int r^{2}d\theta$∫ r² dθ.

The integration methods are in techniques of integration, convergence in sequences and series, and eigenvalues in matrices and Markov chains.