Reverse order of double integral

[Deathfish]

Homework Statement

Integrate (x+2y) over
y=1+x^2 , y=2x^2 and x=0, x=1 (dy dx)

Homework Equations

Graph is sketched.

The Attempt at a Solution

y = 2x^2 --> x=(y/2)^(1/2)
y = 1+x^2 --> x=(y-1)^(1/2)

integrate over y=0 to y=2

problem encountered when solving definite integral from y=0 to y=2, because i end up with (-1/8 y^2 + 1/2 y + 4/5 (y/2)^(1/2) - 4/5 (y-1)^(5/2) - 4/3 (y-1)^(3/2)) over y=0 to y=2, (y-1) gives (-1)^(5/2) and (-1)^(3/2) , square root of negative number.

where did i go wrong?

 

[Ray Vickson]

Homework Statement

Integrate (x+2y) over
y=1+x^2 , y=2x^2 and x=0, x=1 (dy dx)

Homework Equations

Graph is sketched.

The Attempt at a Solution

y = 2x^2 --> x=(y/2)^(1/2)
y = 1+x^2 --> x=(y-1)^(1/2)

integrate over y=0 to y=2

problem encountered when solving definite integral from y=0 to y=2, because i end up with (-1/8 y^2 + 1/2 y + 4/5 (y/2)^(1/2) - 4/5 (y-1)^(5/2) - 4/3 (y-1)^(3/2)) over y=0 to y=2, (y-1) gives (-1)^(5/2) and (-1)^(3/2) , square root of negative number.

where did i go wrong?

Your problem was badly worded, but I guess you mean to find the bivariate integral of (x+2y) over the region bounded by x = 0, x = 1 and lying between the two curves x = 2x^2 and y = 1 + x^2. Is that a correct wording of your problem? If it is, you need to look at two cases: (i) when 2x^2 lies below 1 + x^2; and (ii) when 2x^2 lies above 1 + x^2. Case (i) occurs for some x, while case (ii) occurs for some other x. You need to decide if only one of cases (i) or (ii) will occur for all x in (0,1), or whether both can occur within that x-interval.

RGV

 

[Deathfish]

yeah but i can't figure out how to reverse the order of integration... when i try x=(y/2)^(1/2) to x=(y-1)^(1/2) integrate over y=0 to y=2 i get square root of negative number.

 

[Ray Vickson]

yeah but i can't figure out how to reverse the order of integration... when i try x=(y/2)^(1/2) to x=(y-1)^(1/2) integrate over y=0 to y=2 i get square root of negative number.

Did somebody tell you that you must reverse the order of integration? The easiest way to get the solution is to do the y-integral first (inner integration), then integrate for x from 0 to 1. If you want to revers the order of integration you will have a *more complicated problem*; I suggest you draw a picture of the region and look at what will need to be done for various values of y (that is, what is the x-interval for each possible y). If you draw the picture carefully you will see right away where you went wrong. (In problems of this type it is always a good policy to _draw first, calculate later_!)

RGV

 

[LCKurtz]

yeah but i can't figure out how to reverse the order of integration... when i try x=(y/2)^(1/2) to x=(y-1)^(1/2) integrate over y=0 to y=2 i get square root of negative number.

When you integrate the x variable first you must integrate from x on the left to x on the right. You have to break it up into two pieces because x on the left is not the same for y < 1 that it is when y > 1. Have you drawn a picture?

 

[Deathfish]

Question explicitly states reverse order of integration. I graphed out the region of integration... tried out the obvious x=(y-1)^(1/2) to x=(y/2)^(1/2) and y=0 to y=2, dxdy, but i got a complex number from square root of -1 ... from (0-1)^(1/2) ...

so i tried out various other combinations such as x=0 to x=(y/2)^(1/2) and y=0 to y=1+x^2 ... and worked them out one by one to see if the answer matches the original using an online calculator (save time)... all don't return the same answer. think main problem i have is choosing 4 boundaries when the region graphed has only three surfaces.

 

[jackmell]

You guys mind if I show him first and ask him to justify it or will I get in trouble for that? I think it's effective way to learn it so I'll risk it:

\int_0^b \int_0^{\sqrt{y/2}} f dxdy+\int_b^c \int_{\sqrt{y-1}}^{\sqrt{y/2}} f dxdy

I did it quick so not sure ok. You go over it and either verify it or correct me.

 

[Deathfish]

ok when i draw the region i get only the region to the left of x=1 to x=0, nothing to the right of x=1...

edit : ok is it draw a horizontal line at y=1 ?

 

[LCKurtz]

ok when i draw the region i get only the region to the left of x=1 to x=0, nothing to the right of x=1...

edit : ok is it draw a horizontal line at y=1 ?

Yes. That gives two regions with appropriate, and different, x boundaries.