Integration along y axis

1. Sep 10, 2013

physicsernaw

1. The problem statement, all variables and given/known data
Compute the area as an integral along the y-axis:
f(x) = x^2 - 6, g(x) = 6 - x^3

2. Relevant equations
N/A

3. The attempt at a solution

I solve for x in terms of y for both equations and end up with:
f(y) = +/-√(y+6), g(y) = (6-y)^(1/3)

I then look for interception points of the functions f(y) = g(y)
and I find y = -2.

My question is, if there is only one interception point how can I compute the area
between these two functions? I tried plugging into wolfram, and even it says
"cannot compute integral".

Am I reading the problem wrong? Or am I doing something wrong?

EDIT: When the question states "along the y-axis" does it perhaps mean the line x = 0 as a lower bound?
And then to just integrate from x = 0 to the point of interception (2)? That's the only way I can think of doing this.

Last edited: Sep 10, 2013
2. Sep 11, 2013

abrewmaster

You are right when you say it means to take x=0 as a lower bound, at least that would make the most sense according to the problem statement.

3. Sep 11, 2013

MostlyHarmless

Does the problem say something to the effect of the area bounded by the functions, f(x), g(x), and the y axis?

4. Sep 11, 2013

Zondrina

I find this question peculiar as the graphs $f(y)$ and $g(y)$ will intersect at $(-2,2)$. This creates two different regions ( areas to compute ) in the second quadrant bounded by the line $x=0$.

Is there more to this problem at all?

5. Sep 11, 2013

MostlyHarmless

Yeah, it would be helpful to see EXACTLY what the problem says.

6. Sep 11, 2013

physicsernaw

Verbatim:

"In Exercises 27-44, sketch the region enclosed by the curves and compute its area as an integral along the x- or y-axis."

...

28). y = x^2 - 6, y = 6 - x^3, y-axis

7. Sep 11, 2013

MostlyHarmless

Ok, yes that says the region bounded by those two curves and the y axis, its not saying that you have to integrate with respect to y, however it would make the problem easier.

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