Applications of Integration

(cod: P-77-50-5) Buffon's Needles: Consider a floor made of planks \(L\) centimeters wide. When throwing a needle of length \(l < L\) onto the floor, what is the probability that the needle crosses a joint?

Let \(D\) denote the distance from the center of the needle to the nearest line (joint). Now, consider the normal line passing through the center of the needle, and let \(\theta\) denote the smallest angle formed between the needle and this normal. The figure below illustrates the situation for two distinct throws.

From a throw, we can then observe \(\theta\) and \(D\), where we necessarily have \[0 \leq \theta \leq \pi/2 \ \ \ \text{ and } \ \ \ 0 \leq D \leq L/2.\] That is, the result of a throw can be seen as a random choice of a point \((\theta,D)\) in the rectangle below.

Assume that the probability is uniformly distributed in this rectangle, meaning that the probability \(P_W\) of the point obtained after a throw being in a given region \(W\) of this rectangle is proportional to the area of \(W\). Or equivalently, \[P_W = \dfrac{\text{Area of }W}{\left(\pi L/4\right)},\] where the denominator is the area of the rectangle. Now note that the needle will cross a line exactly when \[D \leq \dfrac{l}{2}\cos(\theta),\] as indicated in the figure below.

Therefore, the needle will cross a line exactly when the pair \((\theta,D)\) is in the pink region indicated in the figure below.

We can then conclude that the probability of the needle crossing a line after a throw is equal to:

\(\dfrac{2(L-l)}{L \pi}\)

\(\dfrac{2l}{L \pi}\)

\(\dfrac{l}{L \pi}\)

\(\dfrac{2(L-l)}{l \pi}\)

\(\dfrac{\pi l}{2L }\)

(cod: P-77-57-8) The objective of this exercise is to calculate the area \(A\) of the region bounded by the ellipse with the equation \[\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1,\] where \(a\) and \(b\) are positive numbers (note that when \(a = b\), the equation defines a circle with radius \(a\)).

By symmetry, we can calculate the area of the region in the first quadrant and multiply it by \(4\).

Select the correct alternative:

We have that \[ A = 4 \int_0^a \dfrac{b}{a}\sqrt{a^2 - x^2} \, dx.\] By performing the trigonometric substitution \begin{eqnarray*} x &=& a \, \tan{\theta} \, \, \, \, \theta \in (-\pi/2,\pi/2), \\ & & \\ dx &=& a \sec^2 \theta \, d \theta, \end{eqnarray*} we obtain \begin{eqnarray*} A &=& 4 \int_0^a \dfrac{b}{a}\sqrt{a^2 - x^2} \, dx \\ & & \\ &=& 4 ab \int_0^{\pi/2} \sec^2{\theta} \, d\theta \\ & & \\ &=& \pi ab. \end{eqnarray*}

We have that \[ A = 4 \int_0^a \dfrac{b}{a}\sqrt{a^2 - x^2} \, dx.\] By performing the trigonometric substitution \begin{eqnarray*} x &=& a \, \cos \, \theta, \, \, \, \theta \in [-\pi/2,\pi/2], \\ & & \\ dx &=& a \, \sin{\theta} \, d \theta, \end{eqnarray*} we obtain \begin{eqnarray*} A &=& 4 \int_0^a \dfrac{b}{a}\sqrt{a^2 - x^2} \, dx \\ & & \\ &=& 2 ab \int_0^{\pi/2} \cos^2{\theta} \, d\theta \\ & & \\ &=& \pi ab. \end{eqnarray*}

We have that \[ A = 4 \int_0^a \dfrac{b}{a}\sqrt{a^2 - x^2} \, dx.\] By performing the trigonometric substitution \begin{eqnarray*} x &=& a \, \cos \, \theta, \, \, \, \theta \in [0,\pi], \\ & & \\ dx &=& - a \, \sin\, {\theta} \, d \theta, \end{eqnarray*} we obtain \begin{eqnarray*} A &=& 4 \int_0^a \dfrac{b}{a}\sqrt{a^2 - x^2} \, dx \\ & & \\ &=& 4 ab \int_0^{\pi/2} \sin^2{\theta} \, d\theta \\ & & \\ &=& \pi ab. \end{eqnarray*}

We have that \[ A = 4 \int_0^a \dfrac{b}{a}\sqrt{a^2 - x^2} \, dx.\] By performing the trigonometric substitution \begin{eqnarray*} x &=& a \, \sin\, \theta, \, \, \, \theta \in [-\pi/2,\pi/2], \\ & & \\ dx &=& a \cos{\theta} \, d \theta, \end{eqnarray*} we obtain \begin{eqnarray*} A &=& 4 \int_0^a \dfrac{b}{a}\sqrt{a^2 - x^2} \, dx \\ & & \\ &=& 4 ab \int_0^{\pi/2} \sin^2{\theta} \, d\theta \\ & & \\ &=& \pi ab. \end{eqnarray*}

We have that \[ A = 4 \int_0^a \dfrac{b}{a}\sqrt{a^2 - x^2} \, dx.\] By performing the trigonometric substitution \begin{eqnarray*} x &=& a \, \sin\, \theta, \, \, \, \theta \in [-\pi/2,\pi/2], \\ & & \\ dx &=& a \cos{\theta} \, d \theta, \end{eqnarray*} we obtain \begin{eqnarray*} A &=& 4 \int_0^a \dfrac{b}{a}\sqrt{a^2 - x^2} \, dx \\ & & \\ &=& 4 ab \int_0^{\pi/2} \cos^2{\theta} \, d\theta \\ & & \\ &=& \pi ab. \end{eqnarray*}

(cod: P-77-51-12) Consider a point \(P = (\cosh{t},\, \sinh\, {t})\) in the first quadrant on the hyperbola \[x^2 - y^2 = 1.\] The area of the pink region in the figure below is equal to:

(cod: P-79-49-12) Consider a flexible and homogeneous cable suspended by its ends, attached to poles, both at the same height. The cable then assumes the shape of a curve known as a catenary (it is worth studying the history of this problem).

The objective of this exercise is to describe an expression for such a curve and study its properties.

Assume that the height of the poles is \(h\) and the distance between the poles is \(2b\). We will position the \(x\) axis at ground level and the \(y\) axis as shown in the figure, assuming that the curve is the graph of a differentiable function \(y = f(x)\) that is even (symmetric with respect to the \(y\) axis).

Let \(P\) denote the lowest point of the curve, as shown in the figure below, and let \(Q = (x,f(x))\) be any point on the curve, with \(x>0\). We will describe the forces acting on the segment of the curve between points \(P\) and \(Q\).

The tension force in the rope is tangent to the curve and can be decomposed into horizontal and vertical components (note that such force is horizontal at point \(P\)). Let \(T(x)\) denote the magnitude of the tension force at \(Q\). If \(\theta = \theta(x)\) is the angle described in the figure, then, since there is no motion, we must have \begin{equation}\label{comp1}T_0 = T(x) \cos(\theta(x)). \end{equation} Note, in particular, that the magnitude of the horizontal component of the tension, given by \(T(x) cos(\theta(x))\), is the same at any point (it does not depend on \(x\)).

From the vertical perspective, we have that \[p(x) = T(x) \sin(\theta(x)),\] where \(p(x)\) is the weight of the cable between \(P\) and \(Q\). Since the cable is homogeneous, let \(\rho\) denote the weight of the cable per unit length and \(L(x)\) the length of the segment of the curve between \(P\) and \(Q\). We then have that \[p(x) = \rho L(x),\] that is, \begin{equation}\label{comp2}\rho L(x) = T(x) \sin(\theta(x)). \end{equation} From (\ref{comp1}) and (\ref{comp2}), we conclude that \[f'(x) = \tan(\theta(x)) = \dfrac{\rho L(x)}{T_0}.\] Considering the constant \[a = \dfrac{\rho}{T_0},\] we have that \[f'(x) = a L(x),\] where the length \(L(x)\) is given by \[L(x) = \int_0^x \sqrt{1 + (f'(t))^2} \, dt.\] By the Fundamental Theorem of Calculus, \(L'(x) = \sqrt{1 + (f'(x))^2}\), thus \begin{eqnarray*} & & f''(x) = a \sqrt{1 + (f'(x))^2} \\ & & \\ &\iff& \dfrac{f''(x)}{\sqrt{1 + (f'(x))^2}} = a. \end{eqnarray*} We then want to obtain a function \(f(x)\) that satisfies the above equation (such an equation is called a second-order differential equation). Writing \(v(x) = f'(x)\), the above differential equation can be written more simply (becoming a first-order equation): \[\dfrac{v'(x)}{\sqrt{1 + (v(x))^2}} = a.\] This equation can be solved by integrating both sides. In the integral that arises on the left side, we can make the substitution \(u = v(x)\), obtaining \[\int\dfrac{1}{{\sqrt{1 + u^2}}} \, du,\] where we observe that the integrand is the derivative of the inverse hyperbolic sine. Therefore, we obtain \begin{eqnarray*} & & \operatorname{arsinh}\, (v(x)) = ax + B \\ & & \\ &\iff& v(x) = \sinh\, (ax+B) \\ & & \\ &\iff& f'(x) = \sinh\, (ax+B). \end{eqnarray*} Consider the following questions:

(i) Use \(f'(0)\) to define the value of the constant \(B\). Then, integrate again to obtain the expression for \(f(x)\). In doing so, a new integration constant will arise. Use the value of \(f(b)\) to determine this constant and define the expression for \(f(x)\).

(ii) Determine the length of this catenary. Knowing that the function is even, the length of the catenary will be twice the length of the segment in the first quadrant (segment with positive \(x\)).

(iii) The expression of the catenary depends on the parameter \(a\). To make this clear, let us denote the catenary by \(y = f_a(x)\) and its length by \(\mathcal{L}_a\). Determine the values of the limits \[\lim\limits_{a \to 0^+} f_a(0) \, \, \, \text{ and } \, \, \, \lim\limits_{a \to 0^+} \mathcal{L}_a.\]

(i) We have that \[f(x) = \dfrac{1}{a} \cosh(ax) + h + \dfrac{\cosh(ab)}{a}.\]

(ii) The length of the catenary is equal to \(\dfrac{2\, \sinh(ab)}{a}\).

(iii) We have that \(\lim\limits_{a \to 0^+} f_a(0) = 0\) and \(\lim\limits_{a \to 0^+} \mathcal{L}_a = 2b\).

(i) We have that \[f(x) = \dfrac{1}{a} \cosh(ax) + h + \dfrac{\cosh(ab)}{a}.\]

(ii) The length of the catenary is equal to \(\dfrac{2\, \sinh(b)}{a}\).

(iii) We have that \(\lim\limits_{a \to 0^+} f_a(0) = h\) and \(\lim\limits_{a \to 0^+} \mathcal{L}_a = 2b\).

(i) We have that \[f(x) = \dfrac{1}{a} \cosh(ax) + h - \dfrac{\sinh(ab)}{a}.\]

(ii) The length of the catenary is equal to \(\dfrac{2b\, \cosh(a)}{a}\).

(iii) We have that \(\lim\limits_{a \to 0^+} f_a(0) = h\) and \(\lim\limits_{a \to 0^+} \mathcal{L}_a = b\).

(i) We have that \[f(x) = \dfrac{1}{a} \cosh(ax) + h - \dfrac{\cosh(ab)}{a}.\]

(ii) The length of the catenary is equal to \(\dfrac{2\, \cosh(ab)}{a}\).

(iii) We have that \(\lim\limits_{a \to 0^+} f_a(0) = 0\) and \(\lim\limits_{a \to 0^+} \mathcal{L}_a = b\).

(i) We have that \[f(x) = \dfrac{1}{a} \cosh(ax) + h - \dfrac{\cosh(ab)}{a}.\]

(ii) The length of the catenary is equal to \(\dfrac{2\, \sinh(ab)}{a}\).

(iii) We have that \(\lim\limits_{a \to 0^+} f_a(0) = h\) and \(\lim\limits_{a \to 0^+} \mathcal{L}_a = 2b\).

(cod: P-79-52-8) Consider the surface of revolution (unbounded) obtained by rotating the graph of the function \[f(x) = \dfrac{1}{x}, \, \, \, x \in [1,+\infty)\] around the x-axis. This surface is called the Gabriel's trumpet.

Let \(\mathcal{R}\) be the unbounded region located below the graph of \(f\) and above the \(x\) axis. Denote by \(W\) the solid obtained by rotating the region \(\mathcal{R}\) around the x-axis, that is, \(W\) is the solid bounded by Gabriel's trumpet.

Using improper integration, we can state that:

The surface area of the trumpet is twice the volume enclosed by it (i.e., the volume of \(W\)).

Both the surface area of the trumpet and the volume enclosed by it (i.e., the volume of \(W\)) are finite.

Both the surface area of the trumpet and the volume enclosed by it (i.e., the volume of \(W\)) are infinite.

The surface area of the trumpet is infinite, while the volume enclosed by it (i.e., the volume of \(W\)) is finite.

The surface area of the trumpet is less than the volume enclosed by it (i.e., the volume of \(W\)).

Site:	DICA Project
Course:	Guest Area
Book:	Applications of Integration

Printed by:	Usuário visitante
Date:	Wednesday, 24 June 2026, 4:19 AM

Applications of Integration

Table of contents