Area between a parabola and line? I learned that the distance between two curves is found but taking the integral of the upper curve subtracted by the integral of the lower curve, being evaluated at the intersecting points.
I have equation: $y = x -1$ and $y^2 = 2x + 6$.
I took the square root of $(2x+6)$ so it can be evaluated at $y = \cdots$. And end up with the final answer $13.333$. Wrong answer I am told. What am I doing wrong?
It wants me to find the area under the $y^2 = 2x + 6$ stopping at where the line $x - 1$ intersects.
 A: If $y^2 = 2x+6$ then $y=\pm\sqrt{2x+6}$.
This intersects the line $y=x-1$ twice.  Which one is the upper curve and which is the lower curve depends on $x$.  If $-3\le x\le 2-\sqrt 5$ then both the upper and lower curves are parts of the parabola.  If $2-\sqrt 5 \le x \le 2+\sqrt 5$ then the line is the lower curve and the upper one is $y= \sqrt{2x+6}$.
But that's the hard way.  The easy way is to interchange the roles of $x$ and $y$.  Then the upper curve is the one where $x$ is bigger, not the one where $y$ is bigger.  And that's the line, not the curve.
The curve is $x = \dfrac{y^2+6}{2}$ and the line is $x=y+1$.
$$
\int_a^b (\text{upper} - \text{lower}) \, dy = \int_a^b \left((y+1)-\dfrac{y^2+6}{2} \right) \, dy
$$
The numbers $a$ and $b$ are the $y$-coordinates of the two points of intersection, so you have to find those.
A: Draw the two curves.  This is essential, at least for me.
We get a rightward-opening parabola and a line. The curves meet at $(-1,-2)$ and $(5,4)$.
Now we have several options:
Option 1: Integrate with respect to $y$. Our "right" curve is $x=y+1$, and our "left" curve is $x=\frac{y^2-6}{2}$. The area is
$$\int_{-2}^4 \left(y+1-\frac{y^2-6}{2}\right)\,dy.$$
Option 2: Interchange the roles of $x$ and $y$. So we are looking for the area between $x^2=2y+6$ and $x=y-1$. Presumably you will have no trouble with that.
Option 3: Integrate with respect to $x$. Your diagram will show that from $x=-3$ to $x=-1$, the top curve is $\sqrt{2x=6}$ and the bottom curve is $y=-\sqrt{2x+6}$. Then from $x=-1$ to $x=5$, the top and bottom curves are $x-1$ and $-\sqrt{2x+6}$. So there are two integrals to evaluate. 
