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Math Help - Shell Method help?

  1. #1
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    Shell Method help?

    Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

    y=x^2+3x10
    y=0


    about the x-axis.

    I solved this problem by first solving the first equation and got [-5,2]. And then I used these numbers and performed the shell method and got my answer.

    The answer I got when I did the problem was 69pi.

    But when I checked the answer in the book. It said it was wrong. Can someone show me the steps on how to solve this problem?
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  2. #2
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    Quote Originally Posted by abilitiesz View Post
    Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

    y=x^2+3x10
    y=0


    about the x-axis.

    I solved this problem by first solving the first equation and got [-5,2]. And then I used these numbers and performed the shell method and got my answer.

    The answer I got when I did the problem was 69pi.

    But when I checked the answer in the book. It said it was wrong. Can someone show me the steps on how to solve this problem?

    You have to integrate the square of the function between -5 and 2 and multiply the result by \pi.
    I got 560.23 \pi, and my warmest congratulations to the sadist that came up with such a horrible exercise: things like this drive people away from maths!

    Tonio
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    Ohh I see. How come I didn't have to multiply by 2pi instead of just pi?
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  4. #4
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    Quote Originally Posted by tonio View Post
    You have to integrate the square of the function between -5 and 2 and multiply the result by \pi.
    I got 560.23 \pi, and my warmest congratulations to the sadist that came up with such a horrible exercise: things like this drive people away from maths!

    Tonio
    Maybe that sadist expected students to do the integration itself using technology (facts like that are rarely seen by students as relevant enough to mention ....)
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    Quote Originally Posted by abilitiesz View Post
    Ohh I see. How come I didn't have to multiply by 2pi instead of just pi?

    Because one of the main theoretical bases of this integration's method is the area of a circle, which is \pi r^2\,,\,\,r= the circle's radius.

    Tonio
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  6. #6
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    Quote Originally Posted by tonio View Post
    Because one of the main theoretical bases of this integration's method is the area of a circle, which is \pi r^2\,,\,\,r= the circle's radius.

    Tonio
    Oh, I see. So to clarify for me, when do I know how to use the shell method or the disc method? It's pretty confusing to me right now.
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    Quote Originally Posted by abilitiesz View Post
    Oh, I see. So to clarify for me, when do I know how to use the shell method or the disc method? It's pretty confusing to me right now.
    Make sure you can visualise the solid itself.

    Then, look at the function you're going to use* in the integration and visualise a single Reimman strip. Imagine the solid formed from revolving the strip round each axis. One way you should see a disc or washer, the other way a cylinder. One of them should fit nicely in the target solid - that tells you which formula to choose.

    * assuming you've decided between y(x) vs. x(y) - but you can take each in turn.
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