Applications of Integration | q002.html

(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*}