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Math Help - application integral word problem

  1. #1
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    application integral word problem

    I need some help on this word problem.

    A water tank is in the shape of a right circular cone of altitude 10 feet and base radius 5 feet, with it's vertex at the ground. If the tank is full, find the work done in pumping all of the water out the top of the tank.

    Note: The weight of water is not given so I assume it is the standard 62.4pi.

    Would I need to slice the interval [0,10] or do I need to factor in the base radius?

    Is this close \frac{62.4\pi}{4}\int(y^2)(10-y)dy
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    Quote Originally Posted by gammaman View Post
    I need some help on this word problem.

    A water tank is in the shape of a right circular cone of altitude 10 feet and base radius 5 feet, with it's vertex at the ground. If the tank is full, find the work done in pumping all of the water out the top of the tank.

    Note: The weight of water is not given so I assume it is the standard 62.4pi.

    Would I need to slice the interval [0,10] or do I need to factor in the base radius?

    Is this close \frac{62.4\pi}{4}\int(y^2)(10-y)dy

    sketch the lines y = 2x and y = -2x starting at the origin, up to the points (5,10) and (-5,10)

    weight of a representative slice is ...

    62.4 \, dV = 62.4 \pi \cdot x^2 \, dy = 62.4 \pi \cdot \frac{y^2}{4} \, dy<br />

    the slice needs to be lifted a distance (10 - y) ... work in raising the slice is

    dW = 62.4 \pi \cdot \frac{y^2}{4}(10 - y) \, dy

    total work to raise all slices ...

    W = 15.6 \pi \int_0^{10} y^2(10 - y) \, dy
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    sketch the lines and starting at the origin, up to the points and
    why are we doing this?
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    Quote Originally Posted by gammaman View Post
    why are we doing this?
    it give you a side view of the cone so that you may determine dV and the limits of integration.
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    Where does the base radius come into play? Also I am confused why (y^2) is. Does it just come from the formula for a cone 1/3pi*r^2h. If so why do my notes say to divide by 4?
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  6. #6
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    Quote Originally Posted by gammaman View Post
    Where does the base radius come into play? Also I am confused why (y^2) is. Does it just come from the formula for a cone 1/3pi*r^2h. If so why do my notes say to divide by 4?
    another "why" for sketching a diagram ...

    a horizontal slice of the cone's liquid is a cylinder with radius x and thickness dy.

    since y = 2x , x = \frac{y}{2}

    dV = \pi x^2 \, dy = \pi \left(\frac{y}{2}\right)^2 \, dy = \frac{\pi}{4} y^2 \, dy
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