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Thread: My volume of revolution is only half what it should be

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    My volume of revolution is only half what it should be

    Hi folks,

    I have been trying this problem over and over. I am sure I make some silly arithmetic error, but I just can't see it!

    The task is to find the volume of the solid formed after a single revolution about the x axis. The attached diagram shows that the area is between the curve $y = 4x - x^2$ and $y = 2x$

    so the volume should be $V = \int_{0}^{2} \pi \times (y_{2} - y_{1})^2 dx$

    $V = \pi \int_{0}^{2} (4x - x^2 - 2x)^2 dx$ = $\pi \int_{0}^{2} (2x - x^2)^2 dx$

    $V = \pi \int_{0}^{2} (4x^2 - 4x^3 + x^4) dx$

    $V = \pi \bigg[ \frac{4}{3} x^3 - x^4 + \frac{1}{5} x^5 \bigg]_{0}^{2} = \pi \bigg[ \frac{32}{3} - \frac{16}{1} + \frac{32}{5} \bigg]$

    $V = \pi \bigg[ \frac{160 - 240 + 96}{15} \bigg] = \frac{16\pi}{15}$

    I also tried calculating the volume produced by revolving the area under $y = 4x - x^2$ and subtracting the volume of the cone (height 2 radius 4) but I get the same answer as above.
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    Last edited by s_ingram; Jul 7th 2017 at 03:33 AM.
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    Re: My volume of revolution is only half what it should be

    washer method ... not disks

    $\displaystyle V = \pi \int_0^2 (4x-x^2)^2 -(2x)^2 \, dx$

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    Re: My volume of revolution is only half what it should be

    Thanks Skeeter - I see what you mean... But forgive my English, what does "washer method not disks" mean?
    I see clearly that $y_{1}^2 - y_{2}^2 \neq y_{1}^2 - 2y_{1}y_{2} + y_{2}^2$ but these symbols don't help me see why my method is wrong. All I see is that they are different and yours matches the answer!
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    Re: My volume of revolution is only half what it should be



    Last edited by skeeter; Jul 7th 2017 at 06:16 AM.
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    Re: My volume of revolution is only half what it should be

    $\pi (y_1-y_2)^2$ represents the area of a single disk with radius $y_1-y_2$.

    $\pi y_1^2 - \pi y_2^2$ represents the difference of areas between two disks, one with radius $y_1$ the other with radius $y_2$. You can think of this as a washer. It is one big circle with a smaller circle removed, or a hole in it.

    washer: 2.a small flat ring made of metal, rubber, or plastic fixed under a nut or the head of a bolt to spread the pressure when tightened or between two joining surfaces as a spacer or seal.

    It is called the Washer Method because the drawing of the area being measured frequently looks like a washer.

    Washer Method in Calculus: Formula & Examples - Video & Lesson Transcript | Study.com

    That video may help.
    Last edited by SlipEternal; Jul 7th 2017 at 05:06 AM.
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    Re: My volume of revolution is only half what it should be

    Hi SlipEternal,

    I am somewhat embarrassed here!

    Surely the area of a single disk of radius $y_{1} - y_{2}$ (where $y_{1} \gt y_{2}$) is the same as the difference in area of two disks, one with radius $y_{1}$ and the other $y_{2}$? It represents the same washer.

    In this example I don't actually have a washer, I have a disk. If you look at my diagram, there is no hole. The area between the curve and line is continuous from x = 0 to x = 2. According to Skeeter's diagram for a disk $V = \int_{a}^{b} \pi [R(x)]^2$ where $R(x) = y_{1} - y_{2}$. This is the method that get's me the wrong answer.
    Last edited by s_ingram; Jul 7th 2017 at 09:41 AM.
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    Re: My volume of revolution is only half what it should be

    Quote Originally Posted by s_ingram View Post
    Hi SlipEternal,

    I am somewhat embarrassed here!

    Surely the area of a single disk of radius $y_{1} - y_{2}$ (where $y_{1} \gt y_{2}$) is the same as the difference in area of two disks, one with radius $y_{1}$ and the other $y_{2}$? It represents the same washer.

    In this example I don't actually have a washer, I have a disk. If you look at my diagram, there is no hole. The area between the curve and line is continuous from x = 0 to x = 2. According to Skeeter's diagram for a disk $V = \int_{a}^{b} \pi [R(x)]^2$ where $R(x) = y_{1} - y_{2}$. This is the method that get's me the wrong answer.
    In your original diagram, you show a line segment between the two curves. Imagine you take that line segment and spin it around the x-axis. You wind up with a giant washer. The top of the washer intersects the curve $y=4-x^2$. The inner circle of the washer intersects the line $y=2x$. The "hole" is the space between the curve $y=2x$ and the x-axis. When you revolve it around the x-axis, there is no solid in that space. It is a hole.
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    Re: My volume of revolution is only half what it should be

    ok, I see what you mean, but if you look along the x-axis from the right towards the left, you will not see a hole, you will see the inside of a cone that represents the projection of the line y = 2x.

    If I took a thin slice along the line x = , for example, then it would be a washer. Skeeter's diagram seems to be saying this. I think this is an interesting question though and I am glad it is on the forum.

    By the way, I was further confused because when I read Skeeter's comment: I thought he said "washer not dish" and I was thinking about a machine that washes the dishes! But a dish washer does not revolve like a washing machine and I thought perhaps Skeeter mean't washing machine and this was in some way connected to a volume of revolution. Anyway, that's a complication we don't need to worry about!
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    Re: My volume of revolution is only half what it should be

    Quote Originally Posted by s_ingram View Post
    By the way, I was further confused because when I read Skeeter's comment: I thought he said "washer not dish" and I was thinking about a machine that washes the dishes! But a dish washer does not revolve like a washing machine and I thought perhaps Skeeter mean't washing machine and this was in some way connected to a volume of revolution. Anyway, that's a complication we don't need to worry about!
    washers & disks are the similar cross-sections of volume, dV, for a solid of revolution.

    washers ... a disk with a hole



    disks ... no hole

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