Question

NEED HELP on Calc III Give Medals

Answer 1

GIven that a triple integral can be expressed as\[\int\limits_{0}^{1}\int\limits_{0}^{1-x ^{2}}\int\limits_{0}^{1-x}f(x,y,z)dydzdx\]

Answer 2

Change order of integration to make five equivalents of this integral

Answer 3

So essentially your limits for your first integral are in terms of y. You can think of them as the bounds y = 0 and y = 1-x. Similarly for z and x. The bounds for z can be thought as z = 0 and z = 1-x^2 and x is bounded by 0 and 1.

What I would suggest doing is seeing if you can rough sketch those boundaries and then try to figure out your different orders from there. Have you tried to sketch anything yet to see what's going on and to see what the boundaries for all your variables are?

Answer 4

I have worked one out

Answer 5

And which one is that?

Answer 6

so I wrote that one possibility is\[\int\limits_{0}^{1}\int\limits_{1-y}^{0}\int\limits_{0}^{1-x ^{2}}fdzdxdy\]

Answer 7

f is short for the integrand

Answer 8

sorry, 1-y and 0 should swape

Answer 9

Alright, so we have one with dx last and dy last, now we can make one with dz last. z is bounded by 0 and 1-x^2, which means z is also bounded from 0 to 1. Now we just need to continue to have x and y vary between the appropriate functions. Can you come up with that one?

Answer 10

What do you think it is first?

Answer 11

Also I think integrating fdzdydx should also count

Answer 12

Well of course, I just was sticking with the idea that so far we've switched out the final variable we integrate with respect to. But we will have fdzdydx order as well.

Answer 13

another can be y from 0 to 1, z from 0 to 2y-y^2 and x from 0 to 1-y integrating fdxdzdy

Answer 14

That one works as well :)

Answer 15

it is hard to put dz last becuase z depend on x to the second order

Answer 16

the diagram on m txtbk shows the region is localized in first octet which means no worry about minus sign

Answer 17

Well, z coming last would just require z to be between 0 and 1. We can say \(x=\sqrt{1-z}\)
We don't need the plus or minus since x would only be between 0 and 1 anyway, which gives us x being between 0 and \(\sqrt{1-z}\) and then we can have y still be between 0 and 1-x. So the order:
\[\int\limits_{0}^{1}\int\limits_{0}^{\sqrt{1-z}}\int\limits_{0}^{1-x}fdydxdz\]
Would you agree with that one?

Answer 18

so 1 is\[\int\limits_{0}^{1}\int\limits_{0}^{1-x}\int\limits_{0}^{1-x ^{2}}fdzdydx\]

Answer 19

Sounds good to me :)

Answer 20

2 is \[\int\limits_{0}^{1}\int\limits_{0}^{1-y}\int\limits_{0}^{1-x ^{2}}fdzdxdy\]

Answer 21

I would agree.

Answer 22

3 is\[\int\limits_{0}^{1}\int\limits_{0}^{2y-y ^{2}}\int\limits_{0}^{1-y}fdxdzdy\]

Answer 23

So far so good :3

Answer 24

4 is\[\int\limits_{0}^{1}\int\limits_{0}^{1-y}\int\limits_{0}^{2y-^{2}}fdzdxdy\]

Answer 25

We already have a dzdxdy, though :)

Answer 26

finally we have 5 which is\[\int\limits_{0}^{1}\int\limits_{0}^{\sqrt{1-z}}\int\limits_{0}^{1-x}fdydxdz\]

Answer 27

I thought it could be a different one cause the limits are diff

Answer 28

you need 6 different orders, not just any 6 different representations of the limits. I need

dxdydz
dxdzdy
dydxdz
dydzdx
dzdxdy
dzdydx

Answer 29

so it says " rewirte in five other orders" so mut be different squence of x,y,z?

Answer 30

Correct

Answer 31

SHOOOOOT!

Answer 32

Well, some of our orders are different already, so let's see what we have
We started with
dydzdx then
dzdxdy we have
dxdzdy
dydxdz
dzdydx

I think that's all.

So we need dxdydz

Answer 33

Another homework Q reads: Wirte five other integrals that have he same value as the given one" this time no worry about limits right?

Answer 34

I don't think it would matter based on how it's phrased.

Answer 35

I mean I could use a trick if we don't care about limits

Answer 36

From Calc I we know that if we change sign, we flip limits

Answer 37

In Calc 3, if we flip limit once, we change sign, and we flip limits in two different places, we don't

Answer 38

I think if we can do this, rewriting becomes very easy

Answer 39

Yeah, that would be very easy to do wouldn't it, haha. Maybe it intends for you to have all 6 orders, but phrased it in a way to make you think you could do other things. I personally would just practice actually writing all 6 orders of integration instead of being sneaky and just flipping signs and such :)

Answer 40

THX!