Double Integral: [(x2) + y)]1/2 - Solution & Fubini Theorem

[quietrain]

Homework Statement

evaluate the double integral

[(x²) + y)]^1/2 dxdy

where dx goes from y^(1/2) to 2
dy goes from 0 to 4

The Attempt at a Solution

i read a few sources that seems to say i have to use trig identities to solve such problems, but i don't seem to be able to do it here since y is present.
i can't find a suitable substitution too.

on a side note, if i want to proof fubini theorem to be true, am i suppose to show that the function [(x^2) + y)]^1/2 is continuous across the rectangle specified by the limits? so how do i show that in this case?

thanks!

 

[spamiam]

Try exchanging the order of integration. The region over which you're integrating is given by

<br /> \sqrt{y} \leq x \leq 2<br />

<br /> 0 \leq y \leq 4<br />

Can you rewrite these inequalities so that you integrate w.r.t. y first?

 

[quietrain]

er, if i want to integrate wrt to y first, i just have to apply fubini theorem right? but how do i proof that that function is continuous in the rectangle specified by the limits?

and also, if i integrate wrt y first i get

integrate 2/3 (x²+4)^3/2 - 2/3(x²)^3/2 dx

but how do i integrate that x term :O

i would probably need a suitable trigo subsitution?

 

[spamiam]

Oh, you can't just switch the order without changing the limits of integration (especially since one of the limits for x is a function of y!). Draw a picture of the region using the inequalities in my first post, and then you should be able to see another way to rewrite the inequalities.

It's pretty easy to show that this function is continuous (depending on what you've already seen), but I'm not sure you need to worry about proving continuity for the moment. Fubini's theorem basically just states that we can use iterated integrals to solve this problem.

 

[quietrain]

oh so the region i want is a y=x² graph?

but the region is enclose by the y=4, y-axis and the y=x² right?

so i have to integrate from 0 to x² dy, and then 0 to 2 dx

then i use the rectangle area to minus the above integral to get the desired region?

 

[quietrain]

is it this?

 

[spamiam]

There's a small mistake in your picture: double-check which part you should be shading. For instance, the point (0,1) shouldn't be in the shaded region since \sqrt{y} \leq x.

 

[quietrain]

oh, so its the other region

so i should just take away the 8 - , then the rest is right?

thanks!

 

[spamiam]

Sounds good! Now you just have to integrate. It should be much easier integrating w.r.t. y first.

 

[quietrain]

thanks!

 

[spamiam]

You're welcome! It's very common to find double integrals that are very difficult to integrate in one order, but quite simple using the other.