Question

Find the area of the region bounded by the hyperbola 25x2 − 4y2 = 100 and the line x = 3. (Using trigonometric substitution)

Answer 1

Answer:

23.92

Step-by-step explanation:

We solve for y:

$25x^2-4y^2=100$

$y=5\sqrt{x^4/4-1}$

use trig substitution:

$x=2secu$

$x^2=4sec^2u$

The derivative of x is:

$dx=2secutanu du$

when x=2 $u=0$

when x=3 $u=sec^-^1\cdot(3/2)$

The area is defined as 2xarea:

The area is the integral of the equation:

$A=2\int\limits^a_b {5\sqrt{x^2/4-1} } \, dx$ for range 0 to sec-1(3/2)

Substitute x=2secu

$A=\int\limits^a_b 10{\sqrt{(4/4)sec^2u-1} } \, dx$

$A=10\int\limits^a_b {\sqrt{sec^2i-1}23secutanu } \, du$

We know that sec²-1 = tan²u

$A=20\int\limits^a_b {tan^2u secu} \, du$

$A=20\int\limits^a_b {(sec^2u-1)secu} \, du$

$A=20[\int\limits^a_b {sec^3u} \, du - \int\limits^a_b {secu} \, du]$

After simplifying

$A=10[secutanu-ln(secu+tanu)]$

For the range

$A=10[(3/2)\sqrt{5} /2-ln(3/2+\sqrt{5}/2)]$

$A=23.92$