[Calculus] - Double Integral

[Artic_counter]

Hi guys, I'm kind of stuck on this exercice. I've tried Fubini's theorem, tried to switch to polar coordinates, etc... but I can't seem to solve this little exercice haha. Your help would be much appreciated!

int(int(x*(y^2+2)^(1/4),y=x^2..2),x=0..sqrt(2))

double integral.png

[alloyus]

Re: [Calculus] - Double Integral

Aren't you just taking the integral of that and then taking the integral of your answer? Why do you have to get all fancy? *Doesn't know math though*

[Artic_counter]

Re: [Calculus] - Double Integral

That's the usual method but I can't seem to be able to integrate (y^2+2)^(1/4) which mean I need to find another way xD

[igotrhythm]

Re: [Calculus] - Double Integral

The first step is to notice that they're both definite integrals. Definite integrals produce numbers when evaluated, which is usually done by finding the antiderivative, setting C to 0 and then evaluating the function at each point, followed by subtraction.

Since dy is on the inside, x is to be treated as a constant for the first integration, and so can be moved through the integral sign. Just look at the inner number now:

x defint (x^2, 2) (y^2 + 2)^1/4 dy

I want to u-substitute here, but I can't figure out what a good u would be, so I'm stuck. Try flipping through your table of antiderivatives (you should have one that came with your book). That, or do some algebra-fu with the integrand. When you get something, the first definite integral should come out as an expression in terms of x, making the second integration a bit more straightforward.

[Tidus810]

Re: [Calculus] - Double Integral

Yeah, you definitely have to approach it from a non-traditional way. Sometimes with trickier stuff like that, I might plug something into an online calculator and try to work backwards from what gets spat out. But seriously, what that first integral comes out to is absurd... (hypergeometric function..?)

[iCeCuBEz v2]

Re: [Calculus] - Double Integral

yea its a hypergeometric function rofl

this can't be integrated using conventional methods of integration -.-

[Reincarnate]

Re: [Calculus] - Double Integral

1.40241865


Steps:

Rewrite/switch order to
(Integral from 0 to 2) (Integral from 0 to sqrt(y)) x*(y^2 + 2)^(1/4) dx dy

Solve:
(Integral from 0 to sqrt(y)) x*(y^2 + 2)^(1/4) dx
= (1/2)*y*(y^2 + 2)^(1/4)

Solve:
(Integral from 0 to 2) (1/2)*y*(y^2 + 2)^(1/4) dy
= (2/5) * 2^(1/4) * (3*3^(1/4)-1)

It helps sometimes to sketch the bounds out so you can see if something better can be used. The reason why keeping it in its original form is hard is because the inner part is a nonelementary integral, so that's the first major hint that you should probably try to rework the bounds/switch the order.

[igotrhythm]

Re: [Calculus] - Double Integral

Not an elementary function? Well then. XD

[Wayward Vagabond]

Re: [Calculus] - Double Integral

the problem is you have 0 and 0^2 hanging from rope swings that sqrt2 and 2 are holding

getting rid of the rope swings might help

[Ohaider]

Re: [Calculus] - Double Integral

agreed rope swings dont even have anything to do with math

[qqwref]

Re: [Calculus] - Double Integral

Solve:
(Integral from 0 to 2) (1/2)*y*(y^2 + 2)^(1/4) dy
= (2/5) * 2^(1/4) * (3*3^(1/4)-1)

Oh I see what you did there.

Bix just showed the calculation part, but the intuition is this: you have something like (y^2 + 2)^(1/4), which you can't easily integrate unless it's multiplied by the derivative of the inside function (y^2 + 2), or 2y. If you have that you can just do a substitution (u = y^2 + 2) and it's no problem. So you want to get another factor of y somewhere. When you switch the order of the integrals, because the bounds on x depend on y, you can get that extra factor of y in there, which makes the problem solvable.

[Artic_counter]

Re: [Calculus] - Double Integral

Ohhhh, I see it now! I should've drawn my bounds from the start. Silly me xD
Didn't think of x=sqrt(y).

Thanks a lot !

[Reincarnate]

Re: [Calculus] - Double Integral

Oh I see what you did there.

Bix just showed the calculation part, but the intuition is this: you have something like (y^2 + 2)^(1/4), which you can't easily integrate unless it's multiplied by the derivative of the inside function (y^2 + 2), or 2y. If you have that you can just do a substitution (u = y^2 + 2) and it's no problem. So you want to get another factor of y somewhere. When you switch the order of the integrals, because the bounds on x depend on y, you can get that extra factor of y in there, which makes the problem solvable.

Yep! Good explanation

[Arkuski]

Re: [Calculus] - Double Integral

lol nontraditional methods... switching the order of integration is pretty traditional