Question

find the area of the region bounded by the hyperbola\(9{x^2} - 4{y^2} = 36\) and the line \(x = 3\)

Short answer

the area of the region bounded by the hyperbola\(9{x^2} - 4{y^2} = 36\)and the line\(x = 3\)is given as:\(\frac{{9\sqrt 5 }}{4} + \ln 8 - 3\ln (3 + \sqrt 5 )\)

Step-by-step solution

Step 1: 

Find the area of the region bounded by the curves

The following diagram shows theregion bounded by the curves\(9{x^2} - 4{y^2} = 36\)and \(x = 3\)

Step 2:

Solve for x and compute the integral

Note that this is symmetric about the x-axis so just integrate above the x-axis and double the result.

\(\begin{aligned}{l}9{x^2} - 4{y^2} &= 36\\9{x^2} - 36 &= 4{y^2}\\{y^2} &= \frac{{9{x^2} - 36}}{4}\\y &= \pm \frac{3}{2}\sqrt {{x^2} - 4} \end{aligned}\)

From the graph, observe that the area bounded between the lines \(x = 2\)and\(x = 3\).

So the limits of integration are \(x = 2\)and\(x = 3\).

Step 3: 

Find the area bounded between the curve

\(\begin{aligned}{l}A &= \int\limits_2^3 {\frac{3}{2}\sqrt {{x^2} - 4} dx} \\A &= \frac{3}{2}\int\limits_2^3 {\sqrt {{x^2} - {2^2}} dx} \\A &= \frac{3}{2}\left( {\frac{x}{2}\sqrt {{x^2} - {2^2}} - 2\ln (x + \sqrt {{x^2} - 4} )} \right)_2^3\\A &= \frac{3}{2}\left( {\left( {\frac{3}{2}\sqrt {{3^2} - {2^2}} - 2\ln (3 + \sqrt {{3^2} - 4} )} \right) - \left( {\frac{2}{2}\sqrt {{2^2} - {2^2}} - 2\ln (2 + \sqrt {{2^2} - 4} )} \right)} \right)\\A &= \frac{3}{2}\left( {\left( {\frac{{3\sqrt 5 }}{2} - 2\ln (3 + \sqrt 5 )} \right) - \left( { - 2\ln (2)} \right)} \right)\\A &= \frac{{9\sqrt 5 }}{2} - 3\ln (3 + \sqrt 5 ) + 3\ln (2)\\A &= \frac{{9\sqrt 5 }}{2} - 3\ln (3 + \sqrt 5 ) + \ln (8)\end{aligned}\)

Therefore, the area bounded between the curves is \(\frac{{9\sqrt 5 }}{4} + \ln 8 - 3\ln (3 + \sqrt 5 )\).