MHB Rickkyredu's question at Yahoo Answers (shell method for solid of revolution)

AI Thread Summary
The discussion focuses on using the Shell Method to calculate the volume of a solid formed by rotating the area between the curves y=x^5 and y=sqrt[5]{x} around the y-axis. The volume of an arbitrary shell is expressed as dV=2πrh dx, with r=x and h=x^(1/5)-x^5. The limits of integration are determined by finding the intersection points of the two curves, which are x=0 and x=1. The final volume is computed through integration, resulting in V=48π/77. This method effectively demonstrates the application of the Shell Method in calculus.
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Here is the question:

Shell method calculus?

Use the Shell Method to compute the volume of the solid obtained by rotating the region in the first quadrant enclosed by the graphs of the functions y=x^{5} and y=sqrt[5]{x} about the y-axis.

Here is a link to the question:

Shell method calculus? - Yahoo! Answers

I have posted a link there to this topic so the OP may find my response.
 
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Hello Rickkyredu,

The first thing I like to do in these problems, is look at a graph of the region to be rotated:

View attachment 607

Next, I like to compute the volume of 1 arbitrary shell:

$\displaystyle dV=2\pi rh\,dx$

where:

$\displaystyle r=x,\,h=x^{\frac{1}{5}}-x^5$

and so we have:

$\displaystyle dV=2\pi x\left(x^{\frac{1}{5}}-x^5 \right)\,dx=2\pi\left(x^{\frac{6}{5}}-x^6 \right)\,dx$

Next, we need to find the limits of integration, i.e., the $x$-cooridnates of the points of intersection for the two curves:

$\displaystyle x^{\frac{1}{5}}=x^5$

$\displaystyle x^{\frac{1}{5}}-x^5=0$

$\displaystyle x^{\frac{1}{5}}\left(1-x^{\frac{4}{5}} \right)=0$

We can see then:

$x=0,\,1$

Finally, we sum up all the shells by integrating:

$\displaystyle V=2\pi\int_0^1 x^{\frac{6}{5}}-x^6\,dx=2\pi\left[\frac{5}{11}x^{\frac{11}{5}}-\frac{1}{7}x^7 \right]_0^1=2\pi\left(\frac{5}{11}-\frac{1}{7} \right)=\frac{48\pi}{77}$
 

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