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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
 one year ago
 one year 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
 one year ago
 one year ago

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NoelGrecoBest ResponseYou've already chosen the best response.0
What are the points of intersection?
 one year ago

yakeygleeBest ResponseYou've already chosen the best response.1
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. \[V=2\pi \int_a^b x (fg) dx\]
 one year ago

monroe17Best ResponseYou've already chosen the best response.0
0 and 1 are the points of intersection.. but I can't seem to figure out the function to evaluate the integral at ;/
 one year ago

NoelGrecoBest ResponseYou've already chosen the best response.0
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.
 one year ago

yakeygleeBest ResponseYou've already chosen the best response.1
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?
 one year ago

yakeygleeBest ResponseYou've already chosen the best response.1
I should have put \(\displaystyle \int_0^1\), but you get the idea.
 one year ago

monroe17Best ResponseYou've already chosen the best response.0
how did you know what f and g were..? do you just assume?
 one year ago

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

monroe17Best ResponseYou've already chosen the best response.0
what defines an upper and lower bound?
 one year ago

NoelGrecoBest ResponseYou've already chosen the best response.0
I think you want the DIFFERENCE between the two functions in the integrand.
 one year ago

yakeygleeBest 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).
 one year ago

NoelGrecoBest ResponseYou've already chosen the best response.0
I looked at both. You have the sum, not the difference.
 one year ago

NoelGrecoBest ResponseYou've already chosen the best response.0
Just make sure you get \[\frac{ 9\Pi }{ 10 }\]
 one year ago

yakeygleeBest ResponseYou've already chosen the best response.1
That is what I meant. \[V=2\pi \int_\color{red}{0}^\color{red}{1} x (3 \sqrt x \color{red}{} 3x^2) dx.\]
 one year ago
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