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Thread: finding volumes

  1. #1
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    finding volumes

    $\displaystyle y = \dfrac{x^2}{25} , x = 3 , y = 0 $

    Find the volume V of this solid.

    rotate around the Y-axis.


    do i use the disk method and use the equation

    $\displaystyle v = \displaystyle\int^3_3 \pi r^2 dx$
    the bottom of the integration symbol is supposed to be a negative 3, not positive, i don't know how to fix it on here.

    Thanks!
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  2. #2
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    i am finding these problems interesting! we donot have them in our curriculum. so please forgive my lack of formal theory. the first equation is that of a parabola with vertex (0,0)
    pointing upwards with a=25/4. the area enclosed is 9/25 using integration of 5/2(sqrt(y)).dy from 0 to 9/25. the 1/2 is there to find half the area of the parabola. so we get the area 9/25. now since we rotate it around y-axis we have a circle with radius 3. so to get the volume we multiply the circumference of the circle and the area obtained which gives (54/25)pi. is the answer correct???
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  3. #3
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    i wouldn't know. im the one asking the question. haha
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  4. #4
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    If you want to find the volume you find the of that solid save yourself a lot of trouble and factor out the pi to the outside of the integral and then square your f(x) to get x^4/625. Now you can pull that 1/625 and put it on the outside of the integral with pi on top. Then integrate x^4 to get x^5/5. Multiply pi/625 by x^5/5 getting you (pi x^5)/3125. Then its just plug and chug. Also It looks like at x=0 y=0 so you are looking at 0.3 on the x axis.
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  5. #5
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    You will be pleased when you learn the shell method.
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  6. #6
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    i have already learned the shell method, you can use the shell method for this?

    so within the integration of 3 and -3, would it be, $\displaystyle 2 \pi (x)(x^2/25)dx$ ?
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  7. #7
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    Yeah. Its just an easier way of doing the same thing. I think its from 0 to 3 also as that equation is an upward opening parabola that isn't shifted to either side so when x=0 you will have y = 0 and that will be the only zero.
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  8. #8
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    got it, thanks!
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  9. #9
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    To put "-3" as the lower limit rather than just 3, put "{" and "}" around -3: $\displaystyle \int_{-3}^3$.

    In general, you use "{" and "}" to group symbols together.
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