Area between two given equations Let $x^2 + y^2$ = 6 and $x = y^2$. What is the area between them?
I solved the two equations in terms of y and I graph it in Geogebra: https://www.geogebra.org/classic/ahnk4hfw

Solving the two equations in terms of x, the graph is:
https://www.geogebra.org/classic/nwznqk3c
When I integrate in terms of y (the graph is the first one), the bounds are -$\sqrt 2$ and $\sqrt 2$. The integrand is $ \sqrt {6-y^2} $ - $y^2$. I got 4.63 as the area.
$$\int_{-\sqrt 2}^{\sqrt 2} \sqrt {6-y^2} - y^2 \, dy = 4.63 $$
Although, I am not sure if I expressed the integral correctly because it seems that I am missing one function: the -$\sqrt {6-y^2} $.
When I follow the second graph, isn't it that the whole region is bounded by the two functions, and hence the area is the circle's area? Or should I only choose the positive square roots and not the negative ones?
My last question is, what is the best approach of finding the integral of this case? Is it by integrating with respect to y or x? Do I only need to take the smaller region?
 A: Using $\int ydx$.
These curves intersect at $P_1(2,\sqrt{2}),$ and $P_2(2,-\sqrt{2})$
$$A=2\left[\int_{0}^{2} \sqrt{x} dx+\int_{2}^{\sqrt{6}} \sqrt{6-x^2} dx\right]$$
$$A=\frac{8\sqrt{2}}{3}+[3\pi-2\sqrt{2}-6 \sin^{-1}\sqrt{2/3}]=4.5356$$
A: $x^2+y^2=6$ and $y^2=x$ intersect at $P_2(2,\sqrt{2}),$ and $P_1(2,-\sqrt{2})$. The area included between them is best given by
$$A=\int_{-\sqrt{2}}^{\sqrt{2}} [\sqrt{6-y^2}-y^2] dy= \left[\frac{y}{2}\sqrt{6-y^2}+3\sinh^{-1}(y/\sqrt{6})-\frac{y^3}{3}\right]_{-\sqrt{2}}^{\sqrt{2}}$$
$$A=\frac{2}{3}[\sqrt{2}+9\sin^{-1}(1/\sqrt{3})]=4.6354$$
Note:Here area projected on $y$-axis is a good idea. So $A=\int_{y_1}^{y_2}[x_1(y)-x_2(y)]dy.$
A: There are actually two possible regions:

Or,

You see, the 'area between circle and parabola' is a bit ambiguous without saying which area specifically. But I'll take a guess that you are referring the orange (i.e: 2nd diagram). For computing the area, the y axis method and x axis method are both equivalents. Let's focus on the area in first quadrant for now because you can find the one in fourth by symmetry,

Using integration with x-axis, the area is given as:
$$ \int_0^2 y_{parabola} dx + \int_2^{\sqrt{6}} y_{circle} dx$$
If you had done integration with $y$,
$$ \int_0^{\sqrt{2} } (x_{circle} - x_{parabola})  dy$$
Perhaps the second integral is easier because you're evaluating over the same bounds.
