Find the number b such that the line y=b divides the region bounded by y=0 and y=4-x^2 into two regions with equal area.
I know how to graph it out but I have no idea what to do after that.
[Math] Find the number b such that the line y=b divides the region bounded by y=0 and y=4-x^2 into two regions with equal area.
calculus
Related Solutions
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 $y_\text{bottom}(x)$.
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.
Related Question
- [Math] How to find the line that splits the area into two equal parts
- Calculus – Finding Line y=b Dividing Region Bounded by Curves into Equal Areas
- [Math] Finding line that divides an area into equal halves.
- [Math] Find the number $b$ such that the line $y = b$ divides the region bounded by the curves $y = 16x^2$ and $y = 25$ into two regions with equal area.
Best Answer
You know how to graph it out and you find it symmetric about $y$-axis. So we can let the problem to right side of $y$-axis. The integrals in intervals $[0,b]$ and $[b,4]$ respect to $y$-axis are $$\int_0^b\sqrt{4-y}\,dy=\int_b^4\sqrt{4-y}\,dy$$ and you can proceeded here.