Changing the Order of Integration for a Double Integral: How Do I Evaluate This?

[Refraction]

Homework Statement

Looks like I'm back with another question already I need to change the order of integration for this double integral and then evaluate it, but I get to a point where I'm not sure what to do.

Homework Equations

$$\int^3_{0} \int^9_{y} \sqrt{x}cos(x) dx dy$$

The Attempt at a Solution

With the changed order of integration it needs two integrals added together, this is what I came up with:

$$\int^3_{0} \int^x_{0} \sqrt{x}cos(x) dy dx + \int^9_{3} \int^3_{0} \sqrt{x}cos(x) dy dx$$

And I planned to work them both out separately, but didn't get too far with the first one:

$$= \int^3_{0} \left[y\sqrt{x}cos(x)\right]^{x}_{0} dx$$

$$= \int^3_{0} x\sqrt{x}cos(x) dx$$

I'm not sure if I've made a mistake getting here, but it looks like I need to integrate $$x\sqrt{x}cos(x)$$ and there doesn't seem to be an easy way to do that at all.

 

[tiny-tim]

Hi Refraction!

Your change of order looks fine.

The only problem is how to integrate x^1/2cosx or x^3/2cosx … I don't know any way of doing that (other than using power series).

 

[Refraction]

That's what I was thinking as well, we've never done anything like that in this class before, and it's only supposed to be a small question so I'm not sure why it's like that.

The only thing I can think of is it maybe meaning to change the order and just leave it like that, but it's worded a bit strangely then. Thanks anyway!

 

[penguin007]

Hi Refraction,

How did you change the order?(there was a y ?)

For the computation of the integral, the only way I can see is with power series too...

 

[Refraction]

The line was x = y in the original question, I just used it as y = x for when the order is reversed (so it's in the first half of the reversed order integral now).

 

[penguin007]

I still don't understand...

 

[Refraction]

Well the area bounded by the lines looks something like this:

So with the reversed order of integration (dy dx) for the first double integral, R1, the inner integral is from y = 0 to y = x, and the outer integral is from x = 0 to x = 3.

 

[penguin007]

I got it. thank you very much!

 

[The Chaz]

Ah, Grasshopper. The student has become the master!

 

[tiny-tim]

Woohoo! ​