Another simple double integral

[WrongMan]

Homework Statement

So i think i got this straight since my last question. let's see :)
So my area of integration is: y=4 ; y=x² and y=(x-2)²
the function is |x-1|
i must integrate with respect to dx first.

The Attempt at a Solution

So i sketched the area (see attatchment graphs should be cross at x=1 sorry for my bad sketch xD) and found that the upper limit for dx is x² (so y^½) and the lower is (x-2)(so y^½-2)²
then noticed that since the function is the absolute value of x-1 i can divide it up into 2 functions
-x+1 (for x<1) and x-1 (for x>1) and with that in mind in my sketch i drew the line x=1 so my integrals are:
∫_y^½-2¹(-x+1) dx (this integral is from y^½-2 to 1)
and ∫₁^y½(x-1)dx (this integral is from 1 to y^½)
and then integrate all this with respect to y from 0 to 4.

 

[Charles Link]

On your dx lower limit, try recomputing it. You need to take the negative sqrt of (x-2)^2. (The x-2 for the left half of the parabola is negative.)

 

[WrongMan]

On your dx lower limit, try recomputing it. You need to take the negative sqrt of (x-2)^2. (The x-2 for the left half of the parabola is negative.)

oh right it is (sqrt(y) + 2) damn my distractions
that is what you meant right?

 

[Charles Link]

oh right it is (sqrt(y) + 2) damn my distractions
that is what you meant right?

No. -sqrt(y)+2 .i.e. $-y^{1/2}+2$ (You also needed a +2, but you need the minus sqrt.)

 

[WrongMan]

No. $-y^{1/2}+2$

say whaaaat? XD

so (x-2)²=y <=> x-2=y^½ <=> x= y^½ +2
no?

 

[Charles Link]

say whaaaat? XD

so (x-2)²=y <=> x-2=y^½ <=> x= y^½ +2
no?

When you take the sqrt, there are are two solutions. You want the left one because your boundary is on the left side of this parabola. The left one is the negative sqrt. e.g. sqrt(4)=+/- 2. In this case you want "-2". When y=4, x-2=-2 ==>> x=0 is the root you are looking for.

 

[WrongMan]

When you take the sqrt, there are are two solutions. You want the left one because your boundary is on the left side of this parabola. The left one is the negative sqrt. e.g. sqrt(4)=+/- 2. In this case you want "-2". When y=4, x-2=-2 ==>> x=0 is the root you are looking for.

ohhhh i get it. thanks!

 

[Ray Vickson]

Homework Statement

So i think i got this straight since my last question. let's see :)
So my area of integration is: y=4 ; y=x² and y=(x-2)²
the function is |x-1|
i must integrate with respect to dx first.

The Attempt at a Solution

So i sketched the area (see attatchment graphs should be cross at x=1 sorry for my bad sketch xD) and found that the upper limit for dx is x² (so y^½) and the lower is (x-2)(so y^½-2)²
then noticed that since the function is the absolute value of x-1 i can divide it up into 2 functions
-x+1 (for x<1) and x-1 (for x>1) and with that in mind in my sketch i drew the line x=1 so my integrals are:
∫_y^½-2¹(-x+1) dx (this integral is from y^½-2 to 1)
and ∫₁^y½(x-1)dx (this integral is from 1 to y^½)
and then integrate all this with respect to y from 0 to 4.

Why must you integrate with respect to x first? Did somebody order you to do that? The problem is a lot easier if you integrate first with respect to y.

 

[WrongMan]

Why must you integrate with respect to x first? Did somebody order you to do that? The problem is a lot easier if you integrate first with respect to y.

This is from the same set of problems from yesterday.
here is what it says:
calculate ∫∫_Df(x,y) dxdy where:
then i get like 10 different values for f(x,y) and D

 

[Ray Vickson]

This is from the same set of problems from yesterday.
here is what it says:
calculate ∫∫_Df(x,y) dxdy where:
then i get like 10 different values for f(x,y) and D

I do not see anywhere a statement to the effect that you must integrate first with respect to x. It is perfectly legitimate to interchange the order of integration (first x, then y, or first y, then x), and the notation $dx \, dy$ has nothing to do with it. I really hope you do not think otherwise.

There may be reasons to do x first, then y, if the point of the exercise is to make you practice setting up integration limits, etc., but if the aim of the exercise is to simply compute an integral, you can use any order you want, and do whichever is easier. You might even learn something by doing the same problem two different ways.