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Math Help - Difficulty with Volume Problem

  1. #1
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    Difficulty with Volume Problem

    The interior surface of the bowl of a goblet is formed by rotating y=3x2, 0x2 around the y-axis. Wine is poured into the goblet so that it fills 45% of its capacity. Measured from the bottom of the bowl, what is the depth of the wine?

    I know the a,b values are 0,2. This is how I set the integral up.
    2pi x Integral((x)(3x^2))
    2pi x integral 3x^4
    Can someone help me solve this problem please?
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  2. #2
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    Quote Originally Posted by radioheadfan View Post
    The interior surface of the bowl of a goblet is formed by rotating y=3x2, 0x2 around the y-axis. Wine is poured into the goblet so that it fills 45% of its capacity. Measured from the bottom of the bowl, what is the depth of the wine?

    I know the a,b values are 0,2. This is how I set the integral up.
    2pi x Integral((x)(3x^2))
    2pi x integral 3x^4
    Can someone help me solve this problem please?
    Well, first, (x)(3x^2) is NOT 3x^4! It is 3x^3. It looks like you are using the "shell" method. Okay, then you are integrating 2\pi x, the circumference of the circle the shell is rotated around, times the height of the shell. That is NOT 3x^2!

    You want wine in the glass, don't you? The interior of the glass is above the parabola. If the wine fills the glass to height "h" (and you see, I hope, that you are being asked how high to fill the glass). You should be integrating 2\pi \int_0^2 x(h- 3x^2) dx. That will give the volume of wine in the glass. You want the glass filled to 45% of capacity. So you need to know "capacity". When x= 2, y= 3(2^2)= 12 so the full glass has volumne 2\pi\int_0^2 x(12- 3x^2)dx. Calculate that, take 45% of it, set it equal to your previous integral and solve for h.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by radioheadfan View Post
    The interior surface of the bowl of a goblet is formed by rotating y=3x2, 0x2 around the y-axis. Wine is poured into the goblet so that it fills 45% of its capacity. Measured from the bottom of the bowl, what is the depth of the wine?

    I know the a,b values are 0,2. This is how I set the integral up.
    2pi x Integral((x)(3x^2)) you have one too many x's here
    2pi x integral 3x^4 should be 3x^3
    Can someone help me solve this problem please?
    It seems like it makes more sense to integrate with respect to y.

    y=3x^2\implies x=\sqrt{y/3}.

    Now the integral (using the disk method w.r.t. y) is \pi\int_0^{12}(\sqrt{y/3})^2\,dy=24\pi

    Note that this is the same answer you'd get with the shell method you started with.

    Now 45% of 24\pi is 54\pi/5

    So you want to solve:

    \pi\int_0^h(\sqrt{y/3})^2\,dy=\frac{\pi}{3}\int_0^h y\,dy=\frac{54\pi}{5}
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    Well, first, (x)(3x^2) is NOT 3x^4! It is 3x^3. It looks like you are using the "shell" method. Okay, then you are integrating 2\pi x, the circumference of the circle the shell is rotated around, times the height of the shell. That is NOT 3x^2!

    You want wine in the glass, don't you? The interior of the glass is above the parabola. If the wine fills the glass to height "h" (and you see, I hope, that you are being asked how high to fill the glass). You should be integrating 2\pi \int_0^2 x(h- 3x^2) dx. That will give the volume of wine in the glass. You want the glass filled to 45% of capacity. So you need to know "capacity". When x= 2, y= 3(2^2)= 12 so the full glass has volumne 2\pi\int_0^2 x(12- 3x^2)dx. Calculate that, take 45% of it, set it equal to your previous integral and solve for h.
    -
    For the first integral I got -12pi*h
    for the second I got 24pi
    I did as you said and solved 33.9292 is 45%, and solved for h, got -2pi.
    The answer came out wrong, not sure where I messed up.
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