# Homework Help: Integration: find area enclosed by 2 functions

1. Oct 4, 2012

### zachem62

1. The problem statement, all variables and given/known data

Find the area of the region enclosed by x=2-y2 and x+y=0 integrating with respect to y.

2. Relevant equations

3. The attempt at a solution

this confuses me because when integrating with respect to y, you get the functions in terms of x and then integrate them as you would. but when you do that, there is no enclosed area. what i mean is when you have y=√(2-x2) and y=-x, there is an intersection at x=-1 but thats it...so there is no enclosement and i don't know how to integrate here...please help!
thanks.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 4, 2012

### cheahchungyin

Did you write the functions wrong? x=2-y^2 is not equal to y=√(2-x^2)

3. Oct 4, 2012

### SammyS

Staff Emeritus
When integrating with respect to x, the integrand is in terms of x.

When integrating with respect to y, the integrand is in terms of y.

One boundary, as it is given, has x as function of y, which is what's needed for integrating w.r.t. y. x=2-y2 .

4. Oct 4, 2012

### zachem62

ok well when you have the functions in terms of y, the equations are x=2-y^2 and x=-y and the graph looks like the picture i have in the first attachment and there is an enclosed area and all you do is integrate it from y=-1 to y=2. that seems straightforward to me.

but when you have the functions in terms of x, the equations are y=√(2-x) and y=-x and the graph looks like the picture i have in the second attachment

here you can see that there is an intersection at x=-2 but there is no enclosed area in the graph....what mistake am i making here?

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5. Oct 4, 2012

### cheahchungyin

because y=+-√(2-x)

6. Oct 4, 2012

### cheahchungyin

when you inverse the function x=2-y^2, the resultant is actually two new functions y=√(2-x) and y=-√(2-x).

7. Oct 4, 2012

### SammyS

Staff Emeritus
OK. I misunderstood.

Solving x = 2 - y2 for y requires using a ± symbol, as cheahchungyin said.

8. Oct 5, 2012

### zachem62

yeah thats it. i figured it out. thanks for your help.