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Could someone give me a little jump start on this problem? I would love a complete solution but anyhelp will do.

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- Thread starter RadiationX
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- #1

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Could someone give me a little jump start on this problem? I would love a complete solution but anyhelp will do.

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Is that from negative infinity to infinity in both the x and y axes?

- #3

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the bounds are the different surfaces

- #4

saltydog

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RadiationX said:

Could someone give me a little jump start on this problem? I would love a complete solution but anyhelp will do.

The plot is a cross-section with y=0. You want the green right?

What about a triple in spherical coordinates?

Rock and roll.

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Yes I need the green part of the graph. SaltyDog, what program did you use to generate that graph?

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saltydog

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RadiationX said:Yes I need the green part of the graph. SaltyDog, what program did you use to generate that graph?

Mathematica. Here, this is the code:

Code:

```
<< Graphics`FilledPlot`
y1[x_] := (1 - x^2)^(1/2);
y2[x_] := 2 x^2 + 2;
y3[x_] := 4 (x^2)^(1/2);
pf1 = Plot[{y1[x], y2[x], y3[x]}, {x, -1, 1}, PlotRange -> {{-4, 4}, {-4, 4}},
AspectRatio -> 1]
FilledPlot[{y2[x], y1[x], y3[x]}, {x, -1, 1}, PlotRange -> {{-4, 4}, {-4, 4}},
AspectRatio -> 1]
```

Of course in Mathematica, you can use real square-roots and exponent symbols. I just used the carrot above so I could cut and paste it here.

However, I'm having problems converting it to spherical coordinates. Maybe you or someone else can do better or perhaps another method can be used.

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saltydog

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[tex]z=2x^2+2y^2+2\tag{1}[/tex]

into spherical coordinates (you get a quadratic which you can then solve for rho using the quadratic formula but you need to know which sign to use). You can do that. Then integrate rho from the sphere which is 1 up to this function you expressed as [itex]\rho(\phi)[/tex]. Theta of course goes from 0 to 2[itex]\pi[/itex]. Now phi only goes from 0 to that cone-shaped function:

[tex]z=4\sqrt{x^2+y^2}[/tex]

Now, you can figure what the max on [itex]\phi[/itex] has to be from that equation right. Can you then post the expression for [itex]\rho(\phi)[/itex] you determined for (1) above as well as the triple integral?

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saltydog

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Hey RadiationX, I'm affraid I suggested a method more difficult that it needs to be. Cylindrical shells is the simplest approach and the integrals can be evaluated analytically. Just need to split it up into two (single) integrals. Talk about over-kill up there. :yuck: Can you do this?

You should tell me, "you know Salty, just for recommending that ridiculously complex triple integral up there, you need to spend a lot of time trying to evaluate it analytically, yea, and no Mathematica neither!" :surprised

Edit: Hey, I get the same answer for both though so that's encouraging.

You should tell me, "you know Salty, just for recommending that ridiculously complex triple integral up there, you need to spend a lot of time trying to evaluate it analytically, yea, and no Mathematica neither!" :surprised

Edit: Hey, I get the same answer for both though so that's encouraging.

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- #9

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thanks Saltydog would you evaluate it a little more analyticlly for me?

hehe

hehe

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since iam high concerning in math ,iam looking forward to provide me with math books

(calculus,linear algebra ,abstract algebra,.probpilities,statistics.....

evry thing which contains a lot of examples and illustrations

just u can guide me what are the best websites and adresses on the net

iam very thankfu

- #11

saltydog

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RadiationX said:thanks Saltydog would you evaluate it a little more analyticlly for me?

hehe

Here it is in spherical coordinates:

The relationship between spherical coordinates and cartesian coordinates is:

[tex]x=\rho Sin(\phi) Cos(\theta);\quad y=\rho Sin(\phi) Sin(\theta);\quad z=\rho Cos(\phi)[/tex]

We must first convert the upper function, [itex]z=2x^2+2y^2+2[/itex] to spherical coordinates. Doing this, we obtain:

[tex]\rho Cos(\phi)=2\left[\rho^2 Sin^2(\phi)\left(Cos^2(\theta)+Sin^2(\theta)\left)+1\right][/tex]

Solving for [itex]\rho[/itex]:

[tex]\rho(\phi)=\frac{Cos(\phi)\pm \sqrt{Cos^2(\phi)-16\sin^2(\phi)}}{4Sin^2(\phi)}[/tex]

I'll leave it to others to show that the positive value does not converge in the interval of [itex]\phi[/itex] we are concerned with. Therefore, this function in spherical coordinates is given as:

[tex]\rho(\phi)=\frac{Cos(\phi)-\sqrt{Cos^2(\phi)-16\sin^2(\phi)}}{4Sin^2(\phi)}[/tex]

So [itex]\rho[/itex] will be integrated from the sphere which is 1 up to this function. Now,[itex]\phi[/itex] will be integrated from zero to the function [itex]z=4\sqrt{x^2+y^2}[/itex]. It's symmetric about the z-axis. So letting y=0 and just pick a number for z, say z=1, we have 1=4x or x=1/4. Thus:

[tex]\phi_{max}=ArcTan(1/4)[/tex]

We can now set up the integral to determine the volume:

[tex]V=\int_0^{\phi_{max}} \int_0^{2\pi} \int_1^{\rho(\phi)} \rho^2 Sin(\phi)d\rho d\theta d\phi[/tex]

Now, how about setting it up in cylindrical shells?

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thanks saltydog

- #13

saltydog

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Here it is in cylindrical shells:

The diagram below is a cross-section of the region. Consider thin-walled cylinders circling the z-axis (actually y-axis as used here) and split the region of integration into two sections: Section (1) is from x=0 to x=the point where the circle intersects the line y=4x (first vertical line). Region (2) is from that point to the point where y=4x intersects the top curve [itex]y=2x^2+2[/tex] (second line).

Considering in general the volume of the i'th cylinder as:

[tex]V_i=\pi r_i^2(f_a(x_i)-f_b(x_i))-\pi r_{i+1}^2(f_a(x_i)-f_b(x_i))[/tex]

with:

[tex]x_i=\frac{r_i+r_{i+1}}{2}[/tex]

and forming the Riemann sum and taking the limit as the partition goes to zero we obtain:

(or just using the formula in the Calculus book)

[tex]V=V_a+V_b[/tex]

in which:

[tex]V_a=2\pi \int_0^a x\left(2x^2+2-\sqrt{1-x^2}\right)dx[/tex]

[tex]V_b=2\pi \int_a^b x\left(2x^2+2-4x\right) dx[/tex]

with:

[tex]a=\sqrt{\frac{1}{17}}[/tex]

[tex]b=1[/tex]

That is:

[tex]

\begin{align*}

V &= \int_0^{\phi_{max}} \int_0^{2\pi} \int_1^{\rho(\phi)} \rho^2 Sin(\phi)d\rho d\theta d\phi \\

&= 2\pi \int_0^a x\left(2x^2+2-\sqrt{1-x^2}\right)dx+2\pi \int_a^b x\left(2x^2+2-4x\right) dx\\

\end{align}

[/tex]

The diagram below is a cross-section of the region. Consider thin-walled cylinders circling the z-axis (actually y-axis as used here) and split the region of integration into two sections: Section (1) is from x=0 to x=the point where the circle intersects the line y=4x (first vertical line). Region (2) is from that point to the point where y=4x intersects the top curve [itex]y=2x^2+2[/tex] (second line).

Considering in general the volume of the i'th cylinder as:

[tex]V_i=\pi r_i^2(f_a(x_i)-f_b(x_i))-\pi r_{i+1}^2(f_a(x_i)-f_b(x_i))[/tex]

with:

[tex]x_i=\frac{r_i+r_{i+1}}{2}[/tex]

and forming the Riemann sum and taking the limit as the partition goes to zero we obtain:

(or just using the formula in the Calculus book)

[tex]V=V_a+V_b[/tex]

in which:

[tex]V_a=2\pi \int_0^a x\left(2x^2+2-\sqrt{1-x^2}\right)dx[/tex]

[tex]V_b=2\pi \int_a^b x\left(2x^2+2-4x\right) dx[/tex]

with:

[tex]a=\sqrt{\frac{1}{17}}[/tex]

[tex]b=1[/tex]

That is:

[tex]

\begin{align*}

V &= \int_0^{\phi_{max}} \int_0^{2\pi} \int_1^{\rho(\phi)} \rho^2 Sin(\phi)d\rho d\theta d\phi \\

&= 2\pi \int_0^a x\left(2x^2+2-\sqrt{1-x^2}\right)dx+2\pi \int_a^b x\left(2x^2+2-4x\right) dx\\

\end{align}

[/tex]

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