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.
Here's your answer! Hope it helps.
Best Answer
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.