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 2 years ago
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 yaxis
 2 years ago
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 yaxis

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NoelGreco
 2 years ago
Best ResponseYou've already chosen the best response.0What are the points of intersection?

yakeyglee
 2 years ago
Best ResponseYou've already chosen the best response.1For 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. \[V=2\pi \int_a^b x (fg) dx\]

monroe17
 2 years ago
Best ResponseYou've already chosen the best response.00 and 1 are the points of intersection.. but I can't seem to figure out the function to evaluate the integral at ;/

NoelGreco
 2 years ago
Best ResponseYou've already chosen the best response.0dw: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.

yakeyglee
 2 years ago
Best ResponseYou've already chosen the best response.1Using 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?

yakeyglee
 2 years ago
Best ResponseYou've already chosen the best response.1I should have put \(\displaystyle \int_0^1\), but you get the idea.

monroe17
 2 years ago
Best ResponseYou've already chosen the best response.0how did you know what f and g were..? do you just assume?

yakeyglee
 2 years ago
Best ResponseYou've already chosen the best response.1They were the equations you were given (\(f\) is the upper bound and \(g\) is the lower bound). Does that make sense?

monroe17
 2 years ago
Best ResponseYou've already chosen the best response.0what defines an upper and lower bound?

NoelGreco
 2 years ago
Best ResponseYou've already chosen the best response.0I think you want the DIFFERENCE between the two functions in the integrand.

yakeyglee
 2 years ago
Best ResponseYou've already chosen the best response.1@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
 2 years ago
Best ResponseYou've already chosen the best response.0I looked at both. You have the sum, not the difference.

NoelGreco
 2 years ago
Best ResponseYou've already chosen the best response.0Just make sure you get \[\frac{ 9\Pi }{ 10 }\]

yakeyglee
 2 years ago
Best ResponseYou've already chosen the best response.1That 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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