Simple area between curves question

[Gauss177]

Homework Statement

Find the area of the region enclosed by the following curves:
f(y)=1-y^2
g(y)=y^2-1

Homework Equations

The Attempt at a Solution

I'm confused by the graph because the region enclosed has positive and negative parts, and I can't determine whether f(y)>g(y), g(y)>f(y), or what. I'm not sure what to integrate here.

Thanks, I appreciate the help.

 

[Hootenanny]

You could use the modulus function;

$$A = \int^{a}_{b} \left\|f(y)\right\| + \left\|g(y)\right\|\;dy$$

Also, note that f(y) = -g(y)

 

[Gib Z]

The Solution to the integral $$\int f(y) g(y) dy$$ is zero, as f(y)=-g(y).

You could find the definite integral of each, absolute valued. Or a simpler method would be to use symmetry arguments to realize the solution is $$2\int^{1}_{-1} 1-y^2 dy$$, which you should be fine with.

 

[d_leet]

The Solution to the integral $$\int f(y) g(y) dy$$ is zero, as f(y)=-g(y).

Are you sure that integral is zero? You might want to recheck your math unless you meant that product to be a sum.

 

[matt grime]

Homework Statement

Find the area of the region enclosed by the following curves:
f(y)=1-y^2
g(y)=y^2-1

Homework Equations

The Attempt at a Solution

I'm confused by the graph because the region enclosed has positive and negative parts, and I can't determine whether f(y)>g(y), g(y)>f(y), or what. I'm not sure what to integrate here.

Thanks, I appreciate the help.

The curves meet at +1 and -1. You need the area between the curves as y goes from +1 to -1. I'm not sure why you can't tell that f(y) =>g(y) in that region: f(0)=1 and g(0)=-1, so it is clear which is 'on top'. The area is just the integral of f(y)-g(y); I don't mind telling you this (we aren't supposed to just hand out answers) since it is just what you were told in class/in the book. It doesn't matter whether f or g are positive or negative individually: you only care about one relative to the other, the actual signs of f and g don't matter.

For any functions f, and g, the area bound between them between a and b is always

$$\int_a^b \max(f(t),g(t)) - \min(f(t),g(t))dt$$

a=-1, b=1 here, and f(t)=>g(t) for all t in [-1,1].