anonymous
  • anonymous
Find the volume formed by rotating the region enclosed by: y= 2sqrt(x) and y= x about the line x= 5?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Jhannybean
  • Jhannybean
|dw:1370599221554:dw|
zepdrix
  • zepdrix
|dw:1370601112630:dw|Gives us,|dw:1370601140069:dw|Something like that right? :o a shell?
Jhannybean
  • Jhannybean
in order to find the limits, you need to equal both equations to eachother and get the critical points. \[\large 2\sqrt{x}=x\]\[\large 2\sqrt{x}-x = 0\]\[\large \sqrt{x}(2-\sqrt{x}) = 0\]\[\large x = 0\]\[\large 2- \sqrt{x} = 0 \]\[\large (\sqrt{x})^2=(2)^2 \implies x = \pm 4 \ \text {only take positive outcomes}\ \therefore x=4\] Using cylindrical shell method \[\large V= \int\limits_{0}^{4}2 \pi x [f(x)] dx \]\[\large x= \text{radius}=5-x\]\[\large f(x) =\text{height}=2\sqrt{x}-x\]\[\large V=2\pi \int\limits_{0}^{4}(5-x)(2 \sqrt{x}-x)dx\] now all you have to do is finish integrating!

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Jhannybean
  • Jhannybean
only take positive outcome , therefore x = 4* limits of integration then are from 0 to 4 :)

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