Another simple double integral

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.)

 

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

 

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.

 

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.

 

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.