anonymous
  • anonymous
Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y=3sqrt(x) and y=3x^2 about the y-axis
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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NoelGreco
  • NoelGreco
What are the points of intersection?
anonymous
  • anonymous
For two functions \(f\) and \(g\) where \(f \ge g\), the volume between \(x=a\) and \(x=b\) by the cylindrical shells method is given by the following formula. You find those bounds by finding the points of intersection of \(f\) and \(g\), presumably. \[V=2\pi \int_a^b x (f-g) dx\]
anonymous
  • anonymous
0 and 1 are the points of intersection.. but I can't seem to figure out the function to evaluate the integral at ;/

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NoelGreco
  • NoelGreco
|dw:1361312140525:dw| Each shell is going to have the shape of a rectangle when it is unfolded from the circular shell shape. You are really looking for the volume of that rectangle. V = length X height X thickness The thickness is going to be dx or dy, the height is the difference between the two functions, and the length is the circumference of each shell.
anonymous
  • anonymous
Using the formula, we have \(f=3\sqrt x\) and \(g=3x^2\). So\[V=2\pi \int_a^b x (3 \sqrt x + 3x^2) dx.\]You can simplify and integrate that, yes?
anonymous
  • anonymous
I should have put \(\displaystyle \int_0^1\), but you get the idea.
anonymous
  • anonymous
how did you know what f and g were..? do you just assume?
anonymous
  • anonymous
They were the equations you were given (\(f\) is the upper bound and \(g\) is the lower bound). Does that make sense?
anonymous
  • anonymous
what defines an upper and lower bound?
NoelGreco
  • NoelGreco
I think you want the DIFFERENCE between the two functions in the integrand.
anonymous
  • anonymous
@NoelGreco, I do. Look at the integral again. @monroe17, the upper bound is the function that defines the top part of the bounded area, and the lower bound defines the lower part of the bounded area (look at their positions in the picture you drew).
NoelGreco
  • NoelGreco
I looked at both. You have the sum, not the difference.
NoelGreco
  • NoelGreco
Just make sure you get \[\frac{ 9\Pi }{ 10 }\]
anonymous
  • anonymous
That is what I meant. \[V=2\pi \int_\color{red}{0}^\color{red}{1} x (3 \sqrt x \color{red}{-} 3x^2) dx.\]

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