Multiple integrals applications of integration

[Liquidxlax]

Homework Statement

Find the mass of the solid from problem 42 if the density is proportional to y

42. Find the volume between the planes z=2x+3y+6 and z=2x+7y+8 and over the triangular vector (0,0) (0,3) and (2,1)

The Attempt at a Solution

The reason I'm asking is because I'm not sure whether this is a triple or a double integral, how to set it up and how to incorporate the vectors. There isn't a single example of this type in my book, just straight forward simple ones and only 3 of them... my prof never mentioned this in the lectures either and we're done this part.

So an example or any general help will do. Thanks in advance

 

[LCKurtz]

I doubt your book talks about a "triangular vector". Rather those points are no doubt the vertices of a triangle in the (x,y) plane. You will use these points to get the equations of the sides of the triangle.

You can do it either as a double or triple integral. As a double integral:

\int\int\int_{z_{lower}}^{z_{upper}} 1\, dz dy dx

where the dydx or dxdy limits (whichever is easiest) are figured out from the triangle.
Or as a double integral

\int\int (z_{upper} - z_{lower}\, dydx

with the same dydx or dxdy limits.

 

[HallsofIvy]

The triangular region in the xy-plane has lower side, the line from (0, 0) to (2, 1), given by y= (1/2)x or x= 2y. The upper side, the line from (0, 3) to (2, 1), is given by y= 1 -x or x= 1- y.
If you choose to have your outer integral with respect to x, it will be from 0 to 1 and the next integral, with respect to y, will be from (1/2)x to 1- x. If you choose to have your outer integral with respect to y, it will be from 0 to 3 and the next integral, with respect to x, will be from 2y to 1- y.

In either case, you "innermost" integral, with respect to z, will be from 2x+3y+6 to 2x+7y+8.

You integrand, of course, will be c/y for some constant, c.

 

[Liquidxlax]

The triangular region in the xy-plane has lower side, the line from (0, 0) to (2, 1), given by y= (1/2)x or x= 2y. The upper side, the line from (0, 3) to (2, 1), is given by y= 1 -x or x= 1- y.
If you choose to have your outer integral with respect to x, it will be from 0 to 1 and the next integral, with respect to y, will be from (1/2)x to 1- x. If you choose to have your outer integral with respect to y, it will be from 0 to 3 and the next integral, with respect to x, will be from 2y to 1- y.

In either case, you "innermost" integral, with respect to z, will be from 2x+3y+6 to 2x+7y+8.

You integrand, of course, will be c/y for some constant, c.

I doubt your book talks about a "triangular vector". Rather those points are no doubt the vertices of a triangle in the (x,y) plane. You will use these points to get the equations of the sides of the triangle.

You can do it either as a double or triple integral. As a double integral:

\int\int\int_{z_{lower}}^{z_{upper}} 1\, dz dy dx

where the dydx or dxdy limits (whichever is easiest) are figured out from the triangle.
Or as a double integral

\int\int (z_{upper} - z_{lower}\, dydx

with the same dydx or dxdy limits.

thank you very much for the help. I did integrate dz first, the part i didn't understand was the coordinates. thank you very much for the help. i will try this when i get home

edit i think it was actually a square and i transcribed the problem wrong. what would i do if it was a square?

 

[LCKurtz]

thank you very much for the help. I did integrate dz first, the part i didn't understand was the coordinates. thank you very much for the help. i will try this when i get home

edit i think it was actually a square and i transcribed the problem wrong. what would i do if it was a square?

If the xy region is a square aligned parallel to the axes, the xy integrals will have constant limits. It the square is tilted relative to the xy axes you will need the equations of the edges and break up the xy integrals accordingly.

Also in my previous post I used an integrand of 1 which gives the volume. You need the density in there if you are calculating mass.

 

[Liquidxlax]

If the xy region is a square aligned parallel to the axes, the xy integrals will have constant limits. It the square is tilted relative to the xy axes you will need the equations of the edges and break up the xy integrals accordingly.

Also in my previous post I used an integrand of 1 which gives the volume. You need the density in there if you are calculating mass.

it states that the density is proportional to y, which I'm unsure how to approach.

 

[LCKurtz]

it states that the density is proportional to y, which I'm unsure how to approach.

If an object has mass density \delta(x,y,z) then its mass is

\int\int\int_V \delta(x,y,z)\, dV

In your case \delta =\lambda y.

And an additional note, if there is a variable density you should always use a triple integral. A double integral may sometimes be used but there are pitfalls.

 

[Liquidxlax]

i get 0 when it should equal 5

 

[LCKurtz]

You have indicated that you meant the area to be a square but haven't told us what square. You also haven't shown your steps. How can we help you pinpoint your mistake when we don't know what the problem is nor what method you used to solve it?

 

[Liquidxlax]

You have indicated that you meant the area to be a square but haven't told us what square. You also haven't shown your steps. How can we help you pinpoint your mistake when we don't know what the problem is nor what method you used to solve it?

lol sorry, but i meant to delete that message because i found my error out.oh and for the intial question the coordinates should be 0,0 0,1 1,1 1,0 so i could just use 1-x and multiply the whole thing by 2?

to hard to write it all out since i don't know latex, but i got 19/3 = M

 

[LCKurtz]

lol sorry, but i meant to delete that message because i found my error out.


oh and for the intial question the coordinates should be 0,0 0,1 1,1 1,0 so i could just use 1-x and multiply the whole thing by 2?

You mean because 1-x divides the square into two equal area halves? That raises two issues. One is why would you want to introduce variable limits when the problem already has constant limits? A more serious objection is that if you are integrating a density function, you must check that the density affects both halves equally. It might not and you have to be sure it does before you try to use a symmetry argument. So don't do it that way. It's the same reason you want to do 3D density problems with a triple integral, not a double integral as I mentioned earlier.

 

[Liquidxlax]

You mean because 1-x divides the square into two equal area halves? That raises two issues. One is why would you want to introduce variable limits when the problem already has constant limits? A more serious objection is that if you are integrating a density function, you must check that the density affects both halves equally. It might not and you have to be sure it does before you try to use a symmetry argument. So don't do it that way. It's the same reason you want to do 3D density problems with a triple integral, not a double integral as I mentioned earlier.

i did do a triple integral, but I'm having a problem incorporating the second half of the square into my limits.

 

[LCKurtz]

i did do a triple integral, but I'm having a problem incorporating the second half of the square into my limits.

So give us a complete statement of the problem you are trying to work and show us what you are trying.

 

[Liquidxlax]

So give us a complete statement of the problem you are trying to work and show us what you are trying.

int(0 to 1 (x))int(1-x to the rest of the square (y)) int((2x+3y+6 to 2x+4y+8 (z))ydzdydx

 

[LCKurtz]

So give us a complete statement of the problem you are trying to work and show us what you are trying.

int(0 to 1 (x))int(1-x to the rest of the square (y)) int((2x+3y+6 to 2x+4y+8 (z))ydzdydx

That is not a statement of the problem. It is apparently your take on a solution to the problem. Please state the problem.

 

[Liquidxlax]

That is not a statement of the problem. It is apparently your take on a solution to the problem. Please state the problem.

sorry, i was just in such a panic that i didn't sort my problem out. I figured out what i did wrong and got the answer

but if you're still interested in the old problem.

find the mass of the volume between z=2x+7y+8 and z=2x+4y+6 enclosed by the verticies (0,0) (1,0) (0,1) (1,1) when the density is proportional to y

what i did wrong was i was trying to turn the square into an equation when you just use the coordinates