calculus
Find the area of the region bounded above by the curve $x^2+y^2=2$ and below by the curve $y=x^2$.
Here's my attempt:
Since $y =x^2$, then $x^2+x^2=2$, which simplifies to give $x=1$ or $x=-1$. Since $y^2=2-x^2$, then $y=\sqrt( 2-x^2)$. Finally, I tried to integrate $\sqrt (2-x^2)-x^2$ and I got area $\pi/2 $ as my answer. Please correct me if I'm wrong, thanks!
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.
When they say
case 1:
When curve $y=f(X)$ lies above the $X$ axis
the area under curve is calculated using integration
- $\text{area} = \int y\,\text dx$ with some limits
They actually mean
case 1:
When curve $y=f(X)$ lies above the $X$ axis
the area between the curve and the $x$-axis is calculated using integration
- $\text{area} = \int y\,\text dx$ with some limits
So when you're integrating you're only getting half the area:
and naturally, you need to multiply that by $2$ to get all of it.
Also, remember that the actual integrand in question here is not $y^2 = 16x$, because that's not of the form $y = f(x)$ as requested from the cases. It's actually $y = 4\sqrt{x}$, which is only the upper half of the parabola I've sketched above. That makes it a bit more obvious why you don't get all of it; when integrating $y = 4\sqrt x$, your expressions have no way of knowing that it is only half of something bigger, and it especially can't know that the other half is the exact mirror image. So it does the best it can and finds the area down to the $x$-axis instead.
Best Answer
$$ \begin{align} \int_{-1}^1\left(\sqrt{2-x^2}-x^2\right)\mathrm{d}x &=2\int_0^1\left(\sqrt{2-x^2}-x^2\right)\mathrm{d}x\\ &=4\int_0^{\pi/4}\cos^2(\theta)\,\mathrm{d}\theta-\frac23\\ &=2\int_0^{\pi/4}(\cos(2\theta)+1)\,\mathrm{d}\theta-\frac23\\[3pt] &=\frac\pi2+\frac13\\ \end{align} $$