Find the area of the region bounded by the hyperbola 25x^2 − 4y^2 = 100 and the line x = 3.

First graph the hyperbola. The vertices of hyperbola are at x= +-2.
The region bounded is from x=2 to x=3 on right side of hyperbola.
Notice we can use symmetry here. The hyperbola is reflected about the x-axis so the area below equals area above. This means we only have to worry about finding area of region from x=2 to x=3 above x-axis, then double it to get total area.

Next we need to set up an integral to find top area. We need a function of x. Solve hyperbola for y.
[tex]y = \frac{5}{2} \sqrt{x^2 - 4} [/tex]

Area = 2*Integral from 2 to 3
[tex]A = 5 \int\limits^3_2 {\sqrt{x^2 -4}} \, dx [/tex]

Now lets focus on finding anti-derivative, then go back and apply limits to get Area.
[tex] \int\limits {\sqrt{x^2 -4} \, dx [/tex]

Step 1: Use trig substitution.
x = 2sec(w)
dx = 2sec(w)tan(w) dw

[tex]\int \sqrt{x^2-4} dx = \int 4 sec(w) tan^2 (w) dw[/tex]

Step 2: Use integration by parts
u = 4tan(w) dv = sec(w) tan(w)
du = 4sec^2 v = sec(w)

[tex]\int 4 sec(w) tan^2 (w) dw = 4sec(w) tan(w) - \int 4 sec(w) sec^2 (w) dw \\ \\ \int 4 sec(w) tan^2 (w) dw = 4sec(w) tan(w) - \int 4 sec(w) (tan^2 (w)+1) dw [/tex]

This gives same integral on both sides, it can be simplified to:
[tex]\int 8 sec(w) tan^2 (w) dw = 4sec(w) tan(w) - \int 4 sec(w) dw \\ \\ \int 4sec(w) tan^2 (w) dw = 2sec(w) tan(w) -2ln(sec(w) + tan(w))[/tex]

Step 3: Sub back in x.
x = 2sec ----> sec = x/2
tan^2 = sec^2 - 1 -----> tan = sqrt(x^2 -4)/2

[tex]\int \sqrt{x^2 - 4} = 2sec tan - 2ln(sec +tan) = \frac{1}{2} x \sqrt{x^2-4} -2ln(\frac{x+\sqrt{x^2-4}}{2})[/tex]

Finally, apply limits and find Area:
[tex]A = 5[\frac{1}{2} (3) \sqrt{3^2-4} -2ln(\frac{3+\sqrt{3^2-4}}{2}) -\frac{1}{2} (2) \sqrt{2^2-4} +2ln(\frac{2+\sqrt{2^2-4}}{2})] \\ \\ A = 5[\frac{3\sqrt{5}}{2} -2ln(\frac{3+\sqrt{5}}{2}) -0 + 0] \\ \\ A = \frac{15 \sqrt{5}}{2} - 10ln(\frac{3+\sqrt{5}}{2})[/tex]

AkhileshT AkhileshT · Answer 2 · 2019-09-07T11:53:56+02:00

The area of the bounded region is [tex]\boxed{\bf \dfrac{15}{2}\sqrt{5}-10ln\left|\dfrac{3+\sqrt{5}}{2}\right|}[/tex].

Further explanation:

The standard form of the hyperbola is [tex]\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1[/tex] where, [tex]x[/tex] is the major axis and [tex]y[/tex] is the minor axis.

The formula of [tex]\int\left(\sqrt{x^{2}-a^{2}}\right)dx[/tex] is as follows:

[tex]\boxed{\int\left(\sqrt{x^{2}-a^{2}}\right)dx=\dfrac{x}{2}\sqrt{x^{2}-a^{2}}-\dfrac{a^{2}}{2}ln\left|x+\sqrt{x^{2}-a^{2}}\right|}[/tex]

Given:

The equation of the hyperbola is [tex]25x^{2}-4y^{2}=100[/tex] and the equation of the line is [tex]x=3[/tex].

Calculation:

Step 1:

First we convert the given equation in the standard equation of hyperbola.

The given equation of hyperbola is [tex]25x^{2}-4y^{2}=100[/tex].

Divide the given equation by [tex]100[/tex].

[tex]\begin{aligned}\dfrac{25x^{2}}{100}-\dfrac{4y^{2}}{100}&=1\\ \dfrac{x^{2}}{4}-\dfrac{y^{2}}{25}&=1\end{aligned}[/tex]

Therefore, the given equation in the standard form of hyperbola is [tex]\dfrac{x^{2}}{4}-\dfrac{y^{2}}{25}&=1[/tex].

Step 2:

Now, we find the vertices of the hyperbola.

On comparing the given equation with standard form of hyperbola equation, We get

[tex]\boxed{\begin{aligned}a&=\pm 2\\b&=\pm 5\end{aligned}}[/tex]

The vertices of the hyperbola is [tex](\pm a,0)[/tex] when major axis is the [tex]x[/tex]-axis.

Therefore, the vertices of the hyperbola is [tex](\pm 2,0)[/tex].

Step 3:

Now, sketch the graph of the hyperbola [tex]\dfrac{x^{2}}{4}-\dfrac{y^{2}}{25}&=1[/tex].

The vertices of the hyperbola is [tex](\pm 2,0)[/tex] and the major axis is [tex]x[/tex]-axis.

Therefore, the graph can be drawn as in the attached Figure 1 (attached in the end).

Step 4:

The hyperbola is reflected about the [tex]x[/tex]-axis so the area below equals area above.

Therefore, the total area is double of the area of the region from [tex]x=2[/tex] to [tex]x=3[/tex].

First we write the given equation of the hyperbola in terms of [tex]y[/tex].

[tex]y=\dfrac{5}{2}\sqrt{x^{2}-4}[/tex]

Now integrate the function [tex]y=\frac{5}{2}\sqrt{x^{2}-4}[/tex] with the limit from [tex]x=2[/tex] to [tex]x=3[/tex] as follows:

[tex]\begin{aligned}\int\limits^3_2{y}dx&=\int\limits^3_2{\left(\sqrt{x^{2}-4}}\right)dx \\&=\dfrac{5}{2}\left[\dfrac{x}{2}\sqrt{x^{2}-4}-\dfrac{4}{2}ln\left|x+\sqrt{x^{2}-4}\right|\right]^{3}_{2}\\&=\dfrac{5}{2}\left[\dfrac{3}{2}\sqrt{9-4}-\dfrac{4}{2}ln\left|3+\sqrt{5}\right|-\left(\dfrac{3}{2}\sqrt{0}-\dfrac{4}{2}ln\left|2+\sqrt{0}\right|\right)\right]\\&=\dfrac{5}{2}\left[\dfrac{3}{2}\sqrt{5}-2ln\left|3+\sqrt{5}\right|+2ln\left|2\right|\right]\end{aligned}[/tex]

Further solve the above equation as follows:

[tex]\begin{aligned}\int\limits^3_{2}ydx&=\dfrac{5}{2}\left[\dfrac{3}{2}\sqrt{5}-2ln\left|3+\sqrt{5}\right|+2ln\left|2\right|\right]\\&=\dfrac{5}{2}\left[\dfrac{3}{2}\sqrt{5}-2ln\left|\dfrac{3+\sqrt{5}}{2}\right|\right]\end{aligned}[/tex]

The required area is the double of the above area.

[tex]\begin{aligned}\int\limits^3_{2}ydx&=2\cdot \dfrac{5}{2}\left[\dfrac{3}{2}\sqrt{5}-2ln\left|\dfrac{3+\sqrt{5}}{2}\right|\right]\\&=5\left[\dfrac{3}{2}\sqrt{5}-2ln\left|\dfrac{3+\sqrt{5}}{2}\right|\right]\\&=\dfrac{15}{2}\sqrt{5}-10ln\left|\dfrac{3+\sqrt{5}}{2}\right|\end{aligned}[/tex]

Therefore, the area of the bounded region is [tex]\boxed{\bf \dfrac{15}{2}\sqrt{5}-10ln\left|\dfrac{3+\sqrt{5}}{2}\right|}[/tex].

Learn more:

1. Learn more about the function is graphed below https://brainly.com/question/9590016

2. Learn more about the symmetry for a function https://brainly.com/question/1286775

3. Learn more about midpoint of the segment https://brainly.com/question/3269852

Answer details:

Grade: High school

Subject: Mathematics

Chapter: Applications of derivatives

Keywords: Area, line, curve, limits, integration, bounded region, vertices, hyperbola, right side, reflected, function, anti-derivative, derivative, integral, shaded region, bounded area , interval.