Question

Find the area of the region bounded by the hyperbola 9x^2 -4y^2 =36 and x=3

Answer 1

Here's the graph: http://www.wolframalpha.com/input/?i=plot+9x%5E2+-4y%5E2+%3D36%2Cx%3D3+over+%5B0%2C5%5D

Start off by rewriting y in terms of x only, so you have
\[y=\pm\frac{1}{4}\sqrt{9x^2-36}\]
The area of the region would then be
\[\int_a^3\left[\frac{1}{4}\sqrt{9x^2-36} - \left(-\frac{1}{4}\sqrt{9x^2-36}\right)\right]dx\]
Find the value of a, then integrate.

Answer 2

Well, that's the hard way.

Rewrite: \(\dfrac{x^{2}}{2^{2}} - \dfrac{y^{2}}{3^{2}} = 1\)

This gives immedaitely the vertices at (-2,0) and (2,0)

By symmetry, find this integral: \(2\cdot\int\limits_{2}^{3}\dfrac{3}{2}\sqrt{x^{2}-4}\;dx = 3\cdot\int\limits_{2}^{3}\sqrt{x^{2}-4}\;dx\)

Answer 3

Sure @tkhunny, your way of finding the value of a is simple (I'm a bit rusty when it comes to conic sections), but the integral is the same.

Answer 4

Failing to recognize the symmetry might get penalty points with some teachers of the calculus. That's all I'm trying to say.

Answer 5

The symmetry IS taken into account, just not as explicitly.. (+√) - (-√) = +2√

Answer 6

I disagree. If I were teaching a lesson on symmetry, I'd mark it wrong.

If it were a lesson on integration, it would be fine.

Nevertheless, it is fine as it stands, either way.