Vector calculus- region-density-mass

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Jozefina Gramatikova
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Homework Statement


https://www.physicsforums.com/attachments/229290
Typeset problem statement: region R defined by 0 ≤ z ≤ 1 − (4x² + y²), 4x² + y² ≤ 1, x ≥ 0, y ≥ 0, with density ρ(x, y, z) = 1 + z; parts (i) sketch the region and (ii) evaluate the mass.

Homework Equations



Formula: density = mass / volume, written in words above and as rho = m/V below, with the fraction bars shown.

The Attempt at a Solution


Handwritten solution sheet for a mass calculation: region 0 ≤ z ≤ 1 − (4x² + y²), density ρ = 1 + z, with a 3D sketch of the first-octant region and its intercepts, and mass m = ρV written below.

I am not sure what equation to use for the volume[/B]
 
Attachments
  • Typeset problem statement: region R defined by 0 ≤ z ≤ 1 − (4x² + y²), 4x² + y² ≤ 1, x ≥ 0, y ≥ 0, with density ρ(x, y, z) = 1 + z; parts (i) sketch the region and (ii) evaluate the mass.
    upload_2018-8-15_16-23-12.png
    12 KB · Views: 930
  • Formula: density = mass / volume, written in words above and as rho = m/V below, with the fraction bars shown.
    upload_2018-8-15_16-24-25.png
    2.9 KB · Views: 727
  • Handwritten solution sheet for a mass calculation: region 0 ≤ z ≤ 1 − (4x² + y²), density ρ = 1 + z, with a 3D sketch of the first-octant region and its intercepts, and mass m = ρV written below.
    39245723_483901692074113_9059923629021593600_n.jpg
    10.4 KB · Views: 633
The attempt that you posted is too small and I cannot click on it to get a better view. But from what I can see, I suggest the following.
1. Understand, from the ranges of x, y and z given by you, the shape of the region you are interested in. In other words, answer the first question and sketch the region. In your graph, you only showed a few isolated points. You did not sketch the region in space.
2. The relevant equation posted by you is not adequate. Density = mass / volume is good only if the density is constant. In your case, the density depends on position (your equation: ρ = 1 + z). Then you must use the differential form:

dm = ρ dV,

and integrate this to get the total mass. It will be a triple integral.
 
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Chandra Prayaga said:
The attempt that you posted is too small and I cannot click on it to get a better view. But from what I can see, I suggest the following.
1. Understand, from the ranges of x, y and z given by you, the shape of the region you are interested in. In other words, answer the first question and sketch the region. In your graph, you only showed a few isolated points. You did not sketch the region in space.
2. The relevant equation posted by you is not adequate. Density = mass / volume is good only if the density is constant. In your case, the density depends on position (your equation: ρ = 1 + z). Then you must use the differential form:

dm = ρ dV,

and integrate this to get the total mass. It will be a triple integral.
Handwritten 3D sketch of a tetrahedron-like region in x, y, z axes with vertices at (-1/2, 0, 0), (0, -1, 0), (0, 1, 0) and (0, 0, 1/2), shaded green with a purple triangle at the base; beside it are the lines x=0, y=±1, y=0, x=±1/2 and crossed-out and corrected mass integrals m = ∭ρ(x,y,z) dV.

I hope you can see the picture better now. I was wondering if I need the whole rectangle as a region or just the purple triangle that I sketched there?
 
Attachments
  • Hand-drawn 3D sketch on a whiteboard of a triangular region in the xz-plane with vertices at (-1/2, 0, 0), (0, 1, 0), (0, -1, 0) and (1/2, 0, 0), shaded green, with a purple triangle inside; written beside it are x=0, y=±1, y=0, x=±1/2 and the mass integral m = double integral over R of rho(x, y, z) dA.
    39391323_441607066351057_5453612862436016128_n.jpg
    8.4 KB · Views: 323
  • Handwritten sketch of a 3D region in the xy-plane with z-axis, marking points (0, -1, 0), (0, 1, 0), (-1/2, 0, 0), (1/2, 0, 0), with equations x=0, y=±1, y=0, x=±1/2 and a struck-out integral m = ∭ρ(x,y,z) dA beside m = ∭ρ(x,y,z) dA.
    39229078_246147629366978_5680765097081831424_n.jpg
    20.8 KB · Views: 306
  • Handwritten 3D sketch of a tetrahedron-like region in x, y, z axes with vertices at (-1/2, 0, 0), (0, -1, 0), (0, 1, 0) and (0, 0, 1/2), shaded green with a purple triangle at the base; beside it are the lines x=0, y=±1, y=0, x=±1/2 and crossed-out and corrected mass integrals m = ∭ρ(x,y,z) dV.
    39177607_1094553904033685_3707489651934625792_n.jpg
    22.6 KB · Views: 297
Last edited:
I guess it is just the triangle, because we have
Two inequalities written in math: x ≥ 0, y ≥ 0

Handwritten notes on a whiteboard: at top, m = triple integral of rho(x, y, z) over a region; below, m = triple integral with limits 0 to 1, 0 to 1, 0 to 1/2 of (1 + z) dx dy dz; at bottom, m = with nothing after it.

Is my final equation ok?
 
Attachments
  • Two inequalities written in math: x ≥ 0, y ≥ 0
    upload_2018-8-15_22-58-21.png
    1.4 KB · Views: 328
  • Handwritten triple integral on a whiteboard: m = triple integral with limits 0 to 1 in each variable of (1 + z) dx dy dz, with m = below it and the top line m = triple integral of rho(x, y, z) partially visible.
    39177658_527569687683464_3169768074417012736_n.jpg
    16.1 KB · Views: 329
  • Handwritten notes on a whiteboard: at top, m = triple integral of rho(x, y, z) over a region; below, m = triple integral with limits 0 to 1, 0 to 1, 0 to 1/2 of (1 + z) dx dy dz; at bottom, m = with nothing after it.
    39177658_527569687683464_3169768074417012736_n.jpg
    7.5 KB · Views: 301
Last edited:
Jozefina Gramatikova said:
I was wondering if I need the whole rectangle as a region or just the purple triangle that I sketched there?
Your sketch is nowhere close to being right. It is a three dimensional solid.
What is the range of z? For some arbitrary z in that range, what does the XY lamina look like?