Integration along y axis

[physicsernaw]

Homework Statement

Compute the area as an integral along the y-axis:
f(x) = x^2 - 6, g(x) = 6 - x^3

Homework Equations

N/A

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.

 

[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.

 

[MostlyHarmless]

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

 

[STEMucator]

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?

 

[MostlyHarmless]

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

 

[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

 

[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.