Multiple integrals

[Prometheus]

I've a few homework questions involving multiple integrals. I think most of my problems come from the limits of the integrals. Pointers would be most welcome. This is the 1st one.

[math]\int \int_D2xydxdy[/math] where D is the bounded region between [math]y=\frac{1}{2}x^2[/math], [math]x+y=4[/math] and the [math]x[/math]-axis.

 

Wasn't sure whether to split the integral or change the order if integration here (but both should work, right?), so i tried both and got very different answers.

 

So changing the order of integration i make the limits:

 

[math]\int_2^{-4} \int_{4-x}^{\frac{1}{2}x^2}2xydydx[/math]

 

For which eventually i get 2752 - seems too big.

 

Splitting the integral i get:

 

[math]\int_2^{8} \int_{\sqrt2y}^{4-y}2xydxdy + \int_0^{2} \int_{\sqrt2y}^{2}2xydxdy[/math]

 

Which eventually leads to [math]\frac{332}{3}[/math]. This is closer to what i expected. Either way obviously something wrong somewhere, so could someone tell if the limits are correct? Aslo, is there any way sketching these functions here, might help my understanding.

 

Cheers.

[Daedalus]

Hi Prometheus. I checked the integral and arrived at the correct answer for both, [math]x[/math] and [math]y[/math], axes. The problems you are having is that you do not have the limits of integration arranged properly when first integrating along the [math]y[/math] axis, and you have the wrong limits for both *edit* integrals when first integrating along the [math]x[/math] axis (the [math]y[/math] limits are good). I suspect that you have a graphing calculator that will at least graph multiple functions in two dimensions. I suggest that you simultaneously graph [math](1/2) x^2[/math] and [math]4-x[/math]. Graphing the bounding region will help you identify the lower and upper bounds for the double integral. The key is that the function used for the lower bound will be less than the other function along the axis of integration.

 

Oh... one more thing to note. Do not forget to add a negative sign to [math]\sqrt{2 y}[/math] when dealing with negative values of [math]x[/math]!!! After all, [math]x = \sqrt{2 y}[/math], which has two solutions for values of [math]y > 0[/math].

 

Edited November 29, 2012 by Daedalus

[Prometheus]

Thanks a lot Daedalus,

 

As i thought, my problem was with the limits.

 

 

I don't have a graphics calculator but they were easy functions to sketch - i don't think i'll ever be able to tackle such problems without a sketch.

 

The problems you are having is that you do not have the limits of integration arranged properly when first integrating along the [math]y[/math] axis...The key is that the function used for the lower bound will be less than the other function along the axis of integration.

 

If i imagine a line running through the plane, vertical in the case of the changed integral then the lower function will be the lower limit, so it will be:

 

[math]\int_2^{-4} \int_{\frac{1}{2}x^2}^{4-x}2xydydx[/math]

 

For which i make it 180. But i'm still not convinced about the limits of the outer integral. Thinking of that same vertical line passing through the plane, if i imagine it passing from left to right it first passes the point -4 then 2. So then the upper limit is the first point encountered?

 

I'll try the split integral later.

[Daedalus]

Thanks a lot Daedalus,

 

As i thought, my problem was with the limits.

 

I don't have a graphics calculator but they were easy functions to sketch - i don't think i'll ever be able to tackle such problems without a sketch.

You're welcome! I always graph the bounding region regardless my confidence in picking the correct functions for my bounds, and it is a good habit to do so.

 

 

If i imagine a line running through the plane, vertical in the case of the changed integral then the lower function will be the lower limit, so it will be:

 

[math]\int_2^{-4} \int_{\frac{1}{2}x^2}^{4-x}2xy \ dy \, dx[/math]

 

For which i make it 180. But i'm still not convinced about the limits of the outer integral. Thinking of that same vertical line passing through the plane, if i imagine it passing from left to right it first passes the point -4 then 2. So then the upper limit is the first point encountered?

 

I'll try the split integral later.

You are correct about the limits for the outer integral not being right. The limits of integration along the [math]x[/math] axis are opposite from what you just posted. The lower bound is -4 because it is less than 2. You should have:

 

[math]\int_{-4}^{2} \int_{\frac{1}{2}x^2}^{4-x} 2xy \ dy \, dx[/math]

 

 

Now for the split integral.

 

[math]\int_2^{8} \int_{\sqrt2y}^{4-y}2xy \ dx \, dy + \int_0^{2} \int_{\sqrt2y}^{2}2xy \ dx \, dy [/math]

The limits of integration for the [math]y[/math] axis in the first integral is correct, but the lower bound for [math]x[/math] axis is incorrect. Refer to my hint about radicals in my first post and look at the graph I posted ; ) The limits of integration for the [math]y[/math] axis in the second integral is also correct, but the lower and upper bounds for the [math]x[/math] axis is wrong. Again, refer to my hint about radicals and look at the graph to see if you can spot the mistake.

Edited November 29, 2012 by Daedalus

[Prometheus]

The key is that the function used for the lower bound will be less than the other function along the axis of integration.

 

Is this always the case? I need to be able to visualise why this is the case. Is it simply so we don't end up with negative answers?

 

Any way I got [math]336[/math] when changing the order of integration. I then split it like this:

 

[math]\int_2^{8} \int_{-\sqrt2y}^{4-y}2xy \ dx \, dy + \int_0^{2} \int_{-\sqrt2y}^{\sqrt2y}2xy \ dx \, dy [/math]

 

But i got around [math]371[/math]. Think i'm getter closer to understanding, hopefully this was just a computational error.

[Daedalus]

Is this always the case? I need to be able to visualise why this is the case. Is it simply so we don't end up with negative answers?

 

Any way I got [math]336[/math] when changing the order of integration. I then split it like this:

 

[math]\int_2^{8} \int_{-\sqrt2y}^{4-y}2xy \ dx \, dy + \int_0^{2} \int_{-\sqrt2y}^{\sqrt2y}2xy \ dx \, dy [/math]

 

But i got around [math]371[/math]. Think i'm getter closer to understanding, hopefully this was just a computational error.

Hi Prometheus! Sorry I haven't responded. I have been busy all day and just now got home. As for the double integral, you have everything written correctly and should have arrived at -180 for the answer. I use Mathematica and worked the problem three different ways.

 

 

And yes, your lower bound will always be less than the upper bound except in some special cases. Just remember that if you reverse the bounds on a single integral you basically switch the sign of the result. However, this isn't always the case with a double / triple integral because you have to consider higher dimensions than just two. For instance, this particular problem has a negative answer because the function [math]z=2xy[/math] falls into the negative [math]z[/math] range for most of the bounding region. Besides arriving at the wrong answer, I must congratulate you for actually setting up the integral correctly. Just be careful when constructing your bounds, and I highly recommend getting a TI-89 because you can check your answers and it will graph in both 2D and 3D.

Edited December 1, 2012 by Daedalus

[Prometheus]

Which is exactly the answer i got the very first time i worked it out. But i was under the impression that it didn't make sense to report an integral a a negative, because areas and volumes are not negative. And i thought just taking the absolute value of whatever you had was considered 'cheating', so i thought i had the limits wrong. Is this just something taught by certain mathematicians? Nevermind, you have greatly helped my understanding anyway. As for the two methods, i think i prefer changing the order of integration.

 

 

On to another one:

 

[math]\int_{0}^{12} \int_{\frac{y}{3}}^{4} y \sqrt{36+x^3} \ dx \, dy[/math]

 

I did think about trying to change the variable, but i don't really know what i'm doing with that. So i tried the substitution [math]U=\sqrt{36+x^3}[/math] which led to [math]\frac{3x^2}{2\sqrt{36+x^3}}[/math] which i than solved by parts giving:

 

[math]\int_{0}^{12}y [x\sqrt{36+x^3} - \frac{9}{2}ln|36+x^3|]_{\frac{y}{3}}^{4} \ dy[/math]

 

After computations i got another negative answer, so had thought it was wrong again - but maybe not...?

[Daedalus]

Which is exactly the answer i got the very first time i worked it out. But i was under the impression that it didn't make sense to report an integral a a negative, because areas and volumes are not negative. And i thought just taking the absolute value of whatever you had was considered 'cheating', so i thought i had the limits wrong. Is this just something taught by certain mathematicians? Nevermind, you have greatly helped my understanding anyway. As for the two methods, i think i prefer changing the order of integration.

Although areas and volumes are generally positive values, integrals represent a lot more than just geometrical entities. For instance, if you have the derivative of a function, then you can integrate the derivative and obtain the original function. Of course you will need an initial value in order to solve for the constant of integration, but the point is that the original function (which is the integral of the derivative) may very well have negative and / or positive values. So you see, it is entirely possible for integrals to generate negative values, and in some cases the integral may be undefined pending the interval and function being integrated.

 

As for the next integral you are having problems with, I will try and make a post later tonight once I get back from a concert I am attending. My friends are playing tonight and I get to go backstage and hang out with them and the other bands.

[Prometheus]

Think i got it. I changed the order of integration to make the sums easier:

 

[math]\int_{0}^{4} \int_{0}^{3x} y \sqrt{36+x^3} \ dy \, dx[/math]

 

I'll do it like this, but just wondering was i even close with my first attempt.

 

 

Hope you enjoyed the concert!

[Daedalus]

The concert was awesome, and I had a great time!!! I went back to my house and grabbed my PC because I am staying with my brother this week. Once I have it set up, I will be able to help you. Just by looking at your latest post, I'm wondering if the bounds along the [math]y[/math] axis is correct. I'm not sure how you arrived at [math][0, \, 3x][/math]. Although I will have my computer set up later tonight, perhaps someone else can help you if time is of the essence.

Edited December 4, 2012 by Daedalus

[Prometheus]

Was pretty sure i had deciding on the limits down - i'll wait to see what you get.

 

No rush though, I handed in a while back, just trying to make sense of it.

 

P.S. How hard are these sorts of questions considered? These questions are part of a stats course, I've no uni/college maths background before doing this so not sure where it sits in the scheme of things.

Edited December 5, 2012 by Prometheus

[Daedalus]

The following integral is set up correctly:

 

[math]\int_{0}^{4} \int_{0}^{3x} y \sqrt{36+x^3} \ dy \, dx[/math]

 

I checked both integrals and the one listed above is the easiest one to work. You should get a positive answer : ) To answer your question, at my college, double integrals are covered in Calculus 3. Of course, we also covered cylindrical, spherical, and rectangular coordinates as well as transformations between coordinate systems.

Edited December 6, 2012 by Daedalus

[Prometheus]

I got something like 758, pretty sure it was the right answer. Also pretty sure i've got this down now, so long as i sketch out the functions first. Just need to practice now.

 

Thanks for all your help - i'll be sure to bug you with more questions later - moving onto differential equations soon, which i'm told are like integration problems, but harder.