Double integral - area between two parabola

Question

I have to calculate the area between parabolae $y^2=10x+25$ and $y^2=-6x+9$

However I have not been able to do so. Here's what it looks like:

What I tried to do:

1.

I tried splitting the area at x=-1. I took that $x\in[-\frac{5}{2}, -1]$ for the first area, and $y \in [-\sqrt{10x+25}, \sqrt{10x+25}]$ and $x\in[-1, \frac{5}{2}]$ and $y \in [-\sqrt{-6x+9}, \sqrt{6x+9}]$. For the first area I get that it's $2\sqrt{15}$, but the second area can't be calculated that simply. WolframAlpha returns a complex value.

Note: I may have made a mistake with the boundaries. I didn't know how to proceed because, if we go from bottom to top to find the boundaries for $y$, we find that it's bounded by the "same" function (parabola).

2.

Next, I tried splitting the area at y=0. That way, I would have $ y \in [-\sqrt{15},0]$ for the first part, and $ y\in [0, \sqrt{15}]$ for the second part.

However, here I have problems expressing $x$ in terms of $y$. If I go from left to right for the area below the x axis, I encounter the part of the red parabola that is below the x axis, but I don't know how to express it.. The most I can get is $x=\frac{y^2-25}{10}$ and that's the whole parabola, not the lower half!

Any help?

Answer 1

You can express $x$ in terms of $y$ and that gives bounds of $x$ as,

$\dfrac{y^2-25}{10} \leq x \leq \dfrac{9-y^2}{6}$

And at intersection of both curves, $y^2 = 15 \implies y = \pm \sqrt{15}$

So the integral to find area can be written as,

$\displaystyle \int_{-\sqrt{15}}^{\sqrt{15}} \int_{(y^2-25)/10}^{(9-y^2)/6} dx \ dy $

If you go in the order $dy \ dx$, you will have to split the integral into two, which is what you first did.

Answer 2

Both your approaches are, conceptually correct. However there are mistakes.

For $1$: The second area can definitely be calculated, quite easily. But your limits are incorrect. The upper limit wouldn't be $\frac 52$, rather it'd be $\frac 32$. It gives: $$I=2\int_{-1}^{\frac 32} \sqrt {9-6x}\ dx$$ which I'm sure you can finish.

For $2$: You should observe that if $y^2=-6x_1+9$ and $y^2=10 x_2+25$ are the two conics, then upon choosing $y$-axis as base, we get area given by the integral: $$A=\int_{-\sqrt 15}^{\sqrt 15} (x_2-x_1)\ dy$$ Now both $x_1, x_2$ can be represented in terms of $y$, hence $A$ can be evaluated.

Answer 3

The intersection points of the parabolas are the points $\left(-1,\pm\sqrt{15}\right)$.

So, the area is$$\int_{-\sqrt{15}}^{\sqrt{15}}\color{blue}{\frac{9-y^2}6}-\color{red}{\frac{y^2-25}{10}}\,\mathrm dy=16\sqrt{\frac53}.$$