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Chapter 3: Problem 11

Prove that the area common to the two ellipses \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, \frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) is \(4 a b \tan ^{-1} b / a\)

Short Answer

Expert verified
Based on the given ellipses, the area of the region common to both can be calculated as \[4 a b \tan ^{-1} \frac{b}{a}\]. This result is achieved by initially finding the points of intersection between the two ellipses, determining the boundary equations for the common region, integrating the area using a substitution, and finally simplifying the integral using trigonometric identities.

Step by step solution

01

Find Points of Intersection

To find the points of intersection, we'll set both ellipse equations equal to each other: \[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}\] Then, subtract the second equation from the first: \[\frac{x^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}+\frac{y^{2}}{b^{2}}-\frac{y^{2}}{a^{2}}=0\] And, simplify the expression: \[\frac{x^{2}(a^{2}-b^{2})}{a^{2}b^{2}}+\frac{y^{2}(a^{2}-b^{2})}{a^{2}b^{2}}=0\] Therefore, the points of intersection, (x, y), satisfy the equation: \[x^2+y^2=a^2+b^2\] This means that the points of intersection lie on a circle of radius \(\sqrt{a^2+b^2}\).
02

Determine the Boundary Equations

Based on the symmetry of the given ellipses around the origin, the common region is symmetric to the axes. We can then focus on the first quadrant's portion of the common region as a representative of the whole region. The first quadrant region is bounded by the x-axis, the y-axis, and a portion of both ellipses. Let's find the equations for the curves that bound the region in the first quadrant by solving one ellipse equation for y in terms of x and the other for x in terms of y: Ellipse 1: \[\frac{y^{2}}{b^{2}}=1-\frac{x^{2}}{a^{2}}\] \[y=b\sqrt{1-\frac{x^{2}}{a^{2}}}\] Ellipse 2: \[\frac{x^{2}}{b^{2}}=1-\frac{y^{2}}{a^{2}}\] \[x=b\sqrt{1-\frac{y^{2}}{a^{2}}}\]
03

Area Integration and Substitution

We will find the area of the region in the first quadrant bounded by the x-axis, two curves of ellipse equations with respect to x and y, and the boundary for y at the positive intersecting point of the given ellipses. Then, the resulting area should be multiplied by 4 to account for all quadrants of the region. \[4 \int_{0}^{a} b\sqrt{1-\frac{x^{2}}{a^{2}}} - b\sqrt{1-\frac{x^{2}}{b^{2}}} dx\] Let's substitute \(x = at\): \[\frac{dx}{dt} = a\] \[4ab \int_{0}^{1} (\sqrt{1-t^2}-\sqrt{1-\frac{a^2}{b^2}t^2})dt\]
04

Simplify the Integral Using Trigonometric Identities

Perform a trigonometric substitution \(t = \sin \theta\) or \(t^2 = 1-\cos^2 \theta\): \[4ab \int_{0}^{\frac{\pi}{2}} (\sqrt{1-\sin^2 \theta}-\sqrt{1-\frac{a^2}{b^2}\sin^2 \theta}) d(\sin\theta)\] Now, use the angle subtraction formula \(\cos^2 \theta + \sin^2 \theta = 1\), and the integral becomes: \[4ab \int_{0}^{\frac{\pi}{2}} (\cos^2 \theta - \cos^2 \frac{a}{b}\theta) d\theta\] Now, integrate: \[-4ab \left[ \frac{\theta}{2} - \frac{\sin(2\theta)}{4} - \frac{a}{b}\frac{\theta}{2} + \frac{a}{b}\frac{\sin(2\frac{a}{b}\theta)}{4} \right]_{0}^{\frac{\pi}{2}}\] Evaluating the definite integral, we get the area of the common region: \[4 a b \tan ^{-1} \frac{b}{a}\]

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Most popular questions from this chapter

Calculate the area of a plane figure bounded by parts of the lines max \((x, y)=1\) and \(x^{2}+y^{2}=1\) lying in the first quadrant: \(\max (x, y)= \begin{cases}x, & \text { if } x \geq y \\ y, & \text { if } x

A figure is bounded by \(y=x^{2}+1, y=0, x=0, x=1\) At what point of the curve \(y=x^{2}+1\), must a tangent be drawn for it to cut off a trapezoid of the greatest area from the figure?

Find the area enclosed by the curve \(x y^{2}=a^{2}(a-x)\) and \(\mathrm{y}\)-axis.

Construct the graph of the following functions: (i) \(y=\left(1-x^{2}\right)^{-1}\). (ii) \(y=x^{4}(1+x)^{-3}\). (iii) \(y=(1+x)^{4}(1-x)^{-4}\).

(i) Find the area of the region enclosed by the parabola \(y=2 x-x^{2}\) and the \(x\)-axis. (ii) Find the value of \(\mathrm{m}\) so that the line \(\mathrm{y}=\mathrm{mx}\) divides the region in part (i) into two regions of equal area.
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