An area is set from the y-axis, the function y=sqrt(x) and y=4. Calculate the volume of the rotational body if the area is permitted to rotate around the y-axis. Thank you.
First I would make a 2D graph, I will try it here but that wont be pretty :)|dw:1389526607444:dw|
I think that is understandable, is it? That shaded are is what we need, after it is rotated around the y-axis
I came up with a solution that looks like the following: \[V= \int\limits_{0}^{4} 2\pi* x* f(x) dx ==> 2\pi \int\limits_{0}^{4} * x * \sqrt{x} dx ==> 2\pi \int\limits_{0}^{4} x ^{\frac{ 3 }{ 2 }} dx\]
\[2\pi \int\limits_{0}^{4} 0.4 * x ^{\frac{ 5 }{ 2 }} = 1206.79 units. \] Is this correct?
I am a bit confused to be honest, it has been long since Ive done these. But I think you have to take into account that y changes from 0 to 4 and x changes from 0 to 16.
Careful, you are integrating to y=4 right? You have your limit of integration going to x=4.
y=x^(1/2) and y=4 4=x^(1/2) 16=x That should be your upper limit. But keep in mind, what you're integrating is not what you're finding the volume of. What that integral will give you is the negative space of a cylinder.
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