Integration: find area enclosed by 2 functions

[zachem62]

Homework Statement

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

Homework Equations

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 that's it...so there is no enclosement and i don't know how to integrate here...please help!
thanks.

 

[cheahchungyin]

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

 

[SammyS]

Homework Statement

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

Homework Equations

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 that's it...so there is no enclosement and i don't know how to integrate here...please help!
thanks.

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 .

 

[zachem62]

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 .

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?

 

[cheahchungyin]

because y=+-√(2-x)

 

[cheahchungyin]

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

 

[SammyS]

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?

OK. I misunderstood.

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

 

[zachem62]

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

yeah that's it. i figured it out. thanks for your help.