# Homework Help: Volume between Surfaces

1. Sep 12, 2016

### Ronaldo95163

Hey guys.
So i've been trying to learn Double Integrals on my own and i'm at Volume between surfaces...so googling some worksheet problems I came across the one and i'm a bit confused.

1. The problem statement, all variables and given/known data

Let U be the solid above z = 0, below z = 4 − y^2, and between the surfaces x = siny − 1 and x = siny +1. Find the volume of U.

2. Relevant equations

3. The attempt at a solution
So what I was thinking was that the surfaces are x = siny − 1 and x = siny +1. Normally with the volume between surfaces you equate both of them and the resulting function is the region for which the volume is found between the region itself and the difference of the two functions...and the double integral is setup from this.

But here the surfaces are defined in terms of x so I was thinking that the surfaces are in the xy plane and below the 3D parabola in the Z plane z = 4-y^2. So is it the Volume between that and the region enclosed between the surfaces x = siny − 1 and x = siny +1?
My biggest issue in going forward is actually visualizing what's going on so I can setup the double integral

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Last edited by a moderator: Sep 13, 2016
2. Sep 12, 2016

### Simon Bridge

Picture the table is the x-y plane and you have a long loaf laid out along one axis, it sits flat on the table, but is humped up in the parabola shape.
now take a sheet of corrugated roofing iron and use it to cut into the loaf twice to make a wiggly slice ... take away everything but the slice.

3. Sep 12, 2016

### LCKurtz

Mod note: Removed the excess bolding, both in the OP and in the copied text.
In your graphic of the two sine functions, plot the two lines $y=2$ and $y=-2$. The area between those and the two sine curves is your xy domain. The height of your top surface is $z = 4-y^2$ so that is your integrand. Integrate it over your xy region. Do you see which order dxdy or dydx is preferred?

Last edited by a moderator: Sep 13, 2016
4. Sep 13, 2016

### Ronaldo95163

Sorry about the bolding...didn't know...just used the default layout.

Thanks btw. I slept over it and I was thinking to use the z=f(x,y) function as my integrating function as you said over the region bounded by the two curves.
But why use between y=2 & y=-2 though? I know that I have to have some definite values to integrate between but how did you come to the conclusion to use those y values.

BTW dxdy would be preferred since y=-2&2 are constants and they go on the outer integral.

5. Sep 13, 2016

### LCKurtz

Because the problem describes the volume above $z=0$ and under $z=4-y^2$. Where is $z=4-y^2 \ge 0$?

6. Sep 15, 2016

### Ronaldo95163

But wouldn't that be assuming that those two planes intersect each other by equating the both of them?

7. Sep 15, 2016

### Simon Bridge

For the surfaces to describe an enclosed volume, they must intersect. Check: do they intersect?

8. Sep 15, 2016

### Ronaldo95163

Ahh yes they do...the parabola closes down on the xy plane. That means I will be integrating f(x,y) over the region bounded by the two sine functions and the lines y=2 and y = -2 with the order of integration being dxdy

9. Sep 16, 2016

### LCKurtz

Yes. Here's a picture for your amusement:

10. Sep 20, 2016

### Ronaldo95163

Thanks alot man!
What did you use to plot that btw?

11. Sep 20, 2016

Maple.