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Math Help - Volume integration problem

  1. #1
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    Volume integration problem

    Hey I need help with this.

    A giant parabolic space dome has a roof described by the equation

    z = 1 - x^2 - y^2

    where z is the height above the ground and x and y are horizontal coordinates all measured in km.

    Calculate the volume of the dome, using cylindrical polar coordinates.

    Im aware that

    x = rcos0
    y = rsin0
    z = z

    So the volume element is dv = rdrd0dz
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  2. #2
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    Quote Originally Posted by BogStandard View Post
    Hey I need help with this.

    A giant parabolic space dome has a roof described by the equation

    z = 1 - x^2 - y^2

    where z is the height above the ground and x and y are horizontal coordinates all measured in km.

    Calculate the volume of the dome, using cylindrical polar coordinates.

    Im aware that

    x = rcos0
    y = rsin0
    z = z

    So the volume element is dv = rdrd0dz
    Let's first convert the equation of the dome into cylindrical coordinates:

    z = 1 - (r\cos\theta)^2 - (r\sin\theta)^2

    z=1-r^2

    To find the volume, we integrate with respect to all three variables. Now, it doesn't technically matter in which order we do this, but since z is known only in terms of r, it's going to be easiest to make the z variable the first integrated. So:

    \int_0^{2\pi}\int_0^1\int_0^{1-r^2}r\,dz\,dr\,d\theta

    \int_0^{2\pi}\int_0^1(1-r^2)r\,dr\,d\theta

    \int_0^{2\pi}\int_0^1(r-r^3)\,dr\,d\theta

    \int_0^{2\pi}[\frac{r^2}{2}-\frac{r^4}{4}]_0^1\,d\theta

    \frac{1}{4}\int_0^{2\pi}d\theta

    \frac{\pi}{2}
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  3. #3
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    Thanks alot!

    There's another two parts to this question, I tried one of them using the answer here but to no avail.

    b) The density of air inside the dome is given by

    p = a(2 - z)

    where a is a constant with units kgkm^-4. Calculate the total mass of air inside the dome.

    c) A monorail arch spans the dome, running above the line y = 0, i.e. along the path z = 1 - x^2. A train with mass m experiences a force, due to gravity plus an applied electric field, of F = bzi - mgk as it travels along the track. The position vector r = xi + yj + zk. By evaluating the line integral

    INTEGRAL: F.dr

    along the path from the base of the dome at x = -1 to the top of the dome, calculate the work done against F in driving a train along this distance.

    Even just some simple guidance would be appreciated.
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  4. #4
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    Quote Originally Posted by BogStandard View Post
    Thanks alot!

    There's another two parts to this question, I tried one of them using the answer here but to no avail.

    b) The density of air inside the dome is given by

    p = a(2 - z)

    where a is a constant with units kgkm^-4. Calculate the total mass of air inside the dome.
    Mass is the product of volume and density. So, when density is variable, just multiply it by the initial function being integrated for finding the volume, like so:

    m=\rho V=\int_0^{2\pi}\int_0^1\int_0^{1-r^2}\rho r\,dz\,dr\,d\theta

    c) A monorail arch spans the dome, running above the line y = 0, i.e. along the path z = 1 - x^2. A train with mass m experiences a force, due to gravity plus an applied electric field, of F = bzi - mgk as it travels along the track. The position vector r = xi + yj + zk. By evaluating the line integral

    INTEGRAL: F.dr

    along the path from the base of the dome at x = -1 to the top of the dome, calculate the work done against F in driving a train along this distance.

    Even just some simple guidance would be appreciated.
    Remember: work = force * distance

    Unfortunately, since force is variable, we're going to have to set it up in line integral form. So, we need to convert the path to vector form:

    z = 1 - x^2

    \textbf{r}=<x,y,z>

    \textbf{r}=<x,0,1-x^2>

    And the force is given by:

    \textbf{F}=<bz,0,mg>

    \textbf{F}=<b(1-x^2),0,mg>

    The work is denoted by:

    W=\int_C\textbf{F}\cdot d\textbf{r}

    Now, we have \textbf{r}, but what the heck is d\textbf{r}? Just take the derivative...

    d\textbf{r}=<1,0,-2x>dx

    So just plug that in...

    W=\int_C\textbf{F}\cdot d\textbf{r}=\int_{-1}^{0}<b(1-x^2),0,mg>\cdot<1,0,-2x>dx

    You should be able to take it from there.
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  5. #5
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    Any chance you can expand on those last two parts, particularly the 2nd one?
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  6. #6
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    Quote Originally Posted by madpanda View Post
    Any chance you can expand on those last two parts, particularly the 2nd one?
    First problem...

    We left off with:

    m=\rho V=\int_0^{2\pi}\int_0^1\int_0^{1-r^2}\rho r\,dz\,dr\,d\theta

    Now, we had been given the value of \rho:

    \rho=a(2 - z)=2a-az

    So, we just plug that in to the integral:

    m=\int_0^{2\pi}\int_0^1\int_0^{1-r^2}(2a-az)r\,dz\,dr\,d\theta=\int_0^{2\pi}\int_0^1\int_0^  {1-r^2}(2ar-arz)\,dz\,dr\,d\theta

    m=\int_0^{2\pi}\int_0^1[2arz-\frac{arz^2}{2}]_0^{1-r^2}\,dr\,d\theta=\int_0^{2\pi}\int_0^1[2ar(1-r^2)-\frac{ar(1-r^2)^2}{2}]\,dr\,d\theta

    m=\frac{1}{2}\int_0^{2\pi}\int_0^1(3ar-2ar^3-ar^5)\,dr\,d\theta=\frac{1}{2}\int_0^{2\pi}[\frac{9ar^2}{6}-\frac{3ar^4}{6}-\frac{ar^6}{6}]_0^1\,d\theta

    m=\frac{5a}{12}\int_0^{2\pi}d\theta=\frac{5a}{12}[\theta]_0^{2\pi}=\frac{5a\pi}{6}

    Second problem...

    We left off with:

    W=\int_C\textbf{F}\cdot d\textbf{r}=\int_{-1}^{0}<b(1-x^2),0,mg>\cdot<1,0,-2x>dx

    So, let's just evaluate that:

    W=\int_{-1}^{0}(b-2mgx-bx^2)dx=[bx-mgx^2-\frac{bx^3}{3}]_{-1}^{0}=b+mg-\frac{b}{3}=mg+\frac{2b}{3}

    Now, according to the first problem:

    m=\frac{5a\pi}{6}

    So let's plug that into our work function:

    W=mg+\frac{2b}{3}=\frac{5ag\pi}{6}+\frac{2b}{3}

    And that's about it.
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