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Thread: Need help with Volume of 2 Curves about x = 4

  1. #1
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    Need help with Volume of 2 Curves about x = 4

    My two equations are y=10ln(x) and y= -x^4 - x + 4. They are bounded below by the x axis. I need to find the volume of the shape obtained by revolving them around the line x=4. I'm not sure what method I should use to solve this, or how I would set it up. Could somebody please help me with this?
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    Re: Need help with Volume of 2 Curves about x = 4

    Quote Originally Posted by TheNerdyGinger View Post
    My two equations are y=10ln(x) and y= -x^4 - x + 4. They are bounded below by the x axis. I need to find the volume of the shape obtained by revolving them around the line x=4. I'm not sure what method I should use to solve this, or how I would set it up. Could somebody please help me with this?
    Please check your check your class-notes and/or your textbook for corrections to this post.
    If you look at this plot, surely you will agree that the question as stated is ridiculous.
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    Re: Need help with Volume of 2 Curves about x = 4

    Have you graphed the region to be rotated? Using the disk method, each "disk", at a value of y, will have radius equal to 4 minus the x value corresponding to that y. For y from 0 up to the y-value where the two curves cross, the radius will be 4- (4- x- x^4)= x^4+ x. From the y-value where the two curves cross, the radius will be 4- (10 ln(x)). The crucial question is "where do those two curves cross?". Can you solve 10ln(x)= 4- x- x^2?
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    Re: Need help with Volume of 2 Curves about x = 4

    Quote Originally Posted by HallsofIvy View Post
    Have you graphed the region to be rotated? Using the disk method, each "disk", at a value of y, will have radius equal to 4 minus the x value corresponding to that y. For y from 0 up to the y-value where the two curves cross, the radius will be 4- (4- x- x^4)= x^4+ x. From the y-value where the two curves cross, the radius will be 4- (10 ln(x)). The crucial question is "where do those two curves cross?". Can you solve 10ln(x)= 4- x- x^2?
    In the plot can the be bounded below by the x-axis?
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    Re: Need help with Volume of 2 Curves about x = 4

    It's just one of those self imposed restrictions I guess.
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    Re: Need help with Volume of 2 Curves about x = 4

    Are you sure it would be the disk method? Because there would be that space in between them since it's rotated around x = 4.
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    Re: Need help with Volume of 2 Curves about x = 4

    I believe the region to be rotated is small triangular-shaped as shown in the attached graph. Use of cylindrical shells w/respect to x ... a calculator is necessary to determine the intersection of both curves and one x-intercept, not to mention calculating the volume.

    $\displaystyle V=2\pi \int_1^{1.1311} (4-x) \cdot 10\ln{x} \, dx + 2\pi \int_{1.1311}^{1.2838} (4-x) \cdot (4-x-x^4) \, dx$
    Attached Thumbnails Attached Thumbnails Need help with Volume of 2 Curves about x = 4-img_1518.png  
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    Re: Need help with Volume of 2 Curves about x = 4

    That one makes more sense. Thank you!
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    Re: Need help with Volume of 2 Curves about x = 4

    Another question: How would you set it up if I were to revolve those same equations around Y=4? Wouldn't I need to rewrite the equations in terms of y?
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    Re: Need help with Volume of 2 Curves about x = 4

    Quote Originally Posted by TheNerdyGinger View Post
    Another question: How would you set it up if I were to revolve those same equations around Y=4? Wouldn't I need to rewrite the equations in terms of y?
    washers w/respect to x

    two integrals again ...

    $\displaystyle V = \pi \int_1^{1.1311} 4^2 - (4-10\ln{x})^2 \, dx + \pi \int_{1.1311}^{1.2838} 4^2 - (4-x-x^4)^2 \, dx$
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    Re: Need help with Volume of 2 Curves about x = 4

    I guess I over thought that one... I'm starting to understand your tagline.

    Thanks again guys!
    Last edited by TheNerdyGinger; Mar 8th 2017 at 10:02 AM.
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