graphic

• Nov 13th 2008, 07:44 AM
Apprentice123
graphic
How do I construct the graph of the solid limited by: $\displaystyle 2y^2=x$ , $\displaystyle \frac{x}{4}+\frac{y}{2}+\frac{z}{4}=1$ , $\displaystyle z=0$ , $\displaystyle y=0$
• Nov 13th 2008, 09:01 AM
shawsend
Hello Apprentice. I don't know how you guys do those triple integrals without plotting them first. Actually I had this when those guys were helping you with it a few days ago but thought it wasn't necessary. It's a little tough to see without interactively rotating it but below is the domain of integration and the other is the three surfaces. The volume is under the blue section. Look carefully at the domain and then it's easy to see that its:

$\displaystyle \int_0^1\int_{2y^2}^{4-2y}\int_0^{4-x-2y} dzdxdy$

Here's the Mathematica code to draw the surfaces. Also, don't get discouraged by the code. Once you get good at it, it only takes a few minutes to code it.

Code:

polys = Graphics3D[{Opacity[0.2],     LightPurple, {Polygon[{{5, 0, 5},         {-5, 0, 5}, {-5, 0, -5},         {5, 0, -5}}]}}]; polys2 = Graphics3D[{Opacity[0.8],     LightPurple, {Polygon[{{5, 5, 0},         {-5, 5, 0}, {-5, -5, 0},         {5, -5, 0}}]}}]; c1 = ContourPlot3D[{2*y^2 == x}, {x, 0, 5},   {y, -2, 2}, {z, 0, 5}] p1 = Plot3D[4*(1 - x/4 - y/2), {x, 0, 4},   {y, 0, 2}, PlotStyle -> {Opacity[0.5]}] p2 = Plot3D[4*(1 - x/4 - y/2), {x, 0, 4},   {y, 0, 2}, PlotStyle -> Blue,   RegionFunction -> Function[{x, y},     x > 2*y^2 && 4*(1 - x/4 - y/2) > 0]] final = Show[{p1, p2, polys, polys2, c1},   BoxRatios -> {1, 1, 1}, AxesLabel ->     {Style["X", 20], Style["Y", 20],     Style["Z", 20]}] domain = Plot[{Sqrt[x/2], 2 - x/2},   {x, 0, 5}] GraphicsGrid[{{domain, final}}]
• Nov 13th 2008, 09:39 AM
Apprentice123
Quote:

Originally Posted by shawsend
Hello Apprentice. I don't know how you guys do those triple integrals without plotting them first. Actually I had this when those guys were helping you with it a few days ago but thought it wasn't necessary. It's a little tough to see without interactively rotating it but below is the domain of integration and the other is the three surfaces. The volume is under the blue section. Look carefully at the domain and then it's easy to see that its:

$\displaystyle \int_0^1\int_{2y^2}^{4-2y}\int_0^{4-x-2y} dzdxdy$

Here's the Mathematica code to draw the surfaces. Also, don't get discouraged by the code. Once you get good at it, it only takes a few minutes to code it.

Code:

polys = Graphics3D[{Opacity[0.2],     LightPurple, {Polygon[{{5, 0, 5},         {-5, 0, 5}, {-5, 0, -5},         {5, 0, -5}}]}}]; polys2 = Graphics3D[{Opacity[0.8],     LightPurple, {Polygon[{{5, 5, 0},         {-5, 5, 0}, {-5, -5, 0},         {5, -5, 0}}]}}]; c1 = ContourPlot3D[{2*y^2 == x}, {x, 0, 5},   {y, -2, 2}, {z, 0, 5}] p1 = Plot3D[4*(1 - x/4 - y/2), {x, 0, 4},   {y, 0, 2}, PlotStyle -> {Opacity[0.5]}] p2 = Plot3D[4*(1 - x/4 - y/2), {x, 0, 4},   {y, 0, 2}, PlotStyle -> Blue,   RegionFunction -> Function[{x, y},     x > 2*y^2 && 4*(1 - x/4 - y/2) > 0]] final = Show[{p1, p2, polys, polys2, c1},   BoxRatios -> {1, 1, 1}, AxesLabel ->     {Style["X", 20], Style["Y", 20],     Style["Z", 20]}] domain = Plot[{Sqrt[x/2], 2 - x/2},   {x, 0, 5}] GraphicsGrid[{{domain, final}}]

Its building in method $\displaystyle dzdydx$ ???
• Nov 13th 2008, 10:12 AM
shawsend
Quote:

Originally Posted by Apprentice123
Its building in method $\displaystyle dzdydx$ ???

Wait a minute Apprentice . . . that code has nothing to do with the integral. I'm just plotting the graphs without even thinking about the integral, and then looking at the graphs to figure out the limits of an integration. I hope I'm not causing confusion by posting complicated code which interferes with the underlying mathematics.

Just plot one thing:

Code:

p2 = Plot3D[4*(1 - x/4 - y/2), {x, 0, 4},     {y, 0, 2}]
Understand that part before you do anything else. You know why I'm using 4(1-x/4-y/2) right? Now just experiment with that one plot, nothing else: add axes, add labels, add plotting regions, add colors, get that one down perfect then move on to the other pieces. If you don't have Mathematica, use whatever software you have but don't try and do it all at once. It's too confusing.