calculus
now I got region question….
The question was "find the area of the region bounded by the hyperbola $9x^{2} – 4y^{2} = 36$ and the line $x = 3$
I drew the graph of $9x^{2} – 4y^{2} = 36$ and $x = 3$ and I got right side is bounded by $x=3$ but since the graph is hyperbola, left side is not bounded by anything. My question is that how I can find the area of the region ?
If you have any idea, could post here ?
Thanks !!
HINT They ask for the area of the yellow region:
The areas would be given by integrals $\int_{x_1}^{x_2} \left(y_\text{top}(x) - y_\text{bottom}(x)\right) \mathrm{d} x$ with appropriate choices of boundaries $x_1$ and $x_2$ and functions $y_\text{top}(x)$ and $y_\text{bottom}(x)$.
Draw a picture. We get a backward opening parabola, with axis the $x$-axis and vertex at $(64,0)$. Note the symmetry about the $x$-axis..
To find the area, we can integrate with respect to $x$ or with respect to $y$. With respect to $y$ is somewhat easier, but with respect to $x$ is natural for the second part, so that's what we will do.
But I will take advantage of symmetry, because I always do. The area of the top half is $$\int_0^{64}\sqrt{64-x}\,dx.$$ An antiderivative is $-\frac{2}{3}(64-x)^{3/2}$. Plug in the endpoints. We get $\frac{1024}{3}$. It is not useful to give a decimal form. Remember, this is the area of the top half of our region.
We want to choose $a$ so that the area of the top half, up to $a$, is $\frac{512}{3}$. Draw a vertical line that you think splits the region into two equal parts. Note that $a$ should be closer to $0$ than to $64$. We want $$\int_0^a \sqrt{64-x}\,dx=\frac{512}{3}.$$ The integral is $$\frac{2}{3}\left(512-(64-a)^{3/2}\right).$$ Set this equal to $\frac{512}{3}$ and solve for $a$. The equation simplifies to $(64-a)^{3/2}=256$, giving $a=64-32\sqrt[3]{2}$.
Remark: The arithmetic would have been somewhat simpler if we calculated the integral from $x=a$ to $x=64$, and set it equal to $\frac{512}{3}$, but we did it the slightly more awkward way because that's what most students would do.
Instead of calculating the area at the beginning, we could set $$\int_0^a (64-x)^{1/2}\,dx=\int_a^{64}(64-x)^{1/2}\,dx$$ and solve for $a$. The arithmetic is about the same.
Best Answer
Solving for $y$ produces:
$$ y =\frac{1}{2} \sqrt{9x^2-36} $$
What we need to find is the area of $y$ from $x=2$ to $x=3$.
Thus we put:
$$A = \frac{1}{2}\int_2^3 \sqrt{9x^2-36} dx$$
Make an "hiperbolic" substitution:
$$3x = 6 \cosh u$$
$$3dx = 6 \sinh u du$$
We get the new limits are $0$ and $b = \cosh^{-1}\frac{3}{2} = \log({\frac{3}{2} +\sqrt{\frac{5}{4}}})$
$$A = 6\int_0^b {{{\sinh }^2}udu} $$
This integral is very similar to the integral of $\sin^2 x$,
$$A = 6\int_0^b {{{\sinh }^2}udu} =6 \left. {\frac{{\sinh \left( {2u} \right) - 2u}}{4}} \right|_0^b$$
Then you have
$$A = 3\frac{{\sinh 2b - 2b}}{2}$$
After a myriad of algebraic steps I end up with
$$A = \frac{9}{4}\sqrt 5 - 6\log \phi $$
Where $$\phi$$ is the golden ratio number.
Note that
$$\eqalign{ & \log 8 - 3\log \left( {3 + \sqrt 5 } \right) = 3\log 2 - 3\log \left( {3 + \sqrt 5 } \right) = \cr & - 3\log \left( {\frac{{3 + \sqrt 5 }}{2}} \right) = - 3\log \left( {1 + \frac{{1 + \sqrt 5 }}{2}} \right) = \cr & - 3\log \left( {1 + \phi } \right) = - 3\log {\phi ^2} = - 6\log \phi \cr} $$
EDIT: Remeber the value is just half the value of the total area.