Shell/disk method help:( Can someone please help me with this? I do not understand how to get the bounds and do the problem. please give an example and explanation.

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1 Answer
Mar 16, 2018

See below.

Explanation:

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This is for the disc method.

From the diagram we can see that the radius is the distance from #y=0# to the curve #y=x^(1/3)#

.i.e. #y=x^(1/3)#

The radius to the line #y=x# is just #y=x#

Each disc for #y=x^(1/3)# will have a volume of:

#pi(x^(1/3))^2*deltax#

Each disc for #y=x# will have a volume of:

#pi(x^2)*deltax)#

We need to subtract the volume generated by #y=x# from the volume generated by #y=x^(1/3)#

Our bounds are #[0 , 1]#, so our integral will be:

Volume = V.

V#=pi int_(0)^(1)((x^(1/3))^2-(x)^2)dx#

V#=pi int_(0)^(1)(x^(2/3)-x^2)dx=pi{[3/5x^(5/3)-1/3x^3]_(0)^(1)}#

V#=pi{[3/5x^(5/3)-1/3x^3]^(1)-[3/5x^(5/3)-1/3x^3]_(0)}#

Plugging in bounds:

V#=pi{[3/5(1)^(5/3)-1/3(1)^3]^(1)-[3/5(0)^(5/3)-1/3(0)^3]_(0)}#

V#=pi{[4/15]^(1)-[0]_(0)}=color(blue)((4pi)/15 "cubic units.")#

This is for the shell method:

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In this method instead of using disc perpendicular to the axis of rotation, we use cylinders or shells these can be expanded or contracted to completely cover the solid we are seeking. In this method the width of the shell is not the function value as the radius was in the disc method. Here the width of the shell is the distance from the #y# axis. See diagram.

Since we are working in relation to the y axis we need our function to be terms of #y#

i.e. #y=x^(1/3)->x=y^3#

#y=x->x=y#

The distance out to the edge of the shell is #x=1#, so the width of the shell is #1-y^3# see diagram.

The cross sectional area is:

#A(y)=2pi"(radius)(width)"#

#A(y)=2pi(y)(1-y^3)#

#A(y)=2pi(y-y^4)#

The first shell will enter the solid at #y=0# and the final shell at #y=1# ( This is the y value where the two functions intersect) see diagram:

The volume of the solid will then be:

#2pi int_(0)^(1)(y-y^4)dy#

Exactly the same method is used with #x=y#. Remember we need to subtract this volume as in the previous method of the discs.

So our combined integral will be:

#2pi int_(0)^(1)(y-y^4)-(y-y^2)dy#

#2pi int_(0)^(1)(-y^4+y^2)dy=2pi{[-1/5y^5+1/3y^3]_(0)^(1)}#

V#=2pi{[-1/5y^5+1/3y^3]^(1)-[-1/5y^5+1/3y^3]_(0)}#

Plugging in bounds:

V#=2pi{[-1/5(1)^5+1/3(1)^3]^(1)-[-1/5(0)^5+1/3(0)^3]_(0)}#

V#=2pi{[2/15]^(1)-[0]_(0)}=color(blue)((4pi)/15 "cubic units")#

So both methods yield the same result.

Volume of revolution:

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Hope this helps:

As you can see the shell method isn't as intuitive as the disc method. This method takes a fair amount of study to get comfortable with. The main use of this method is when it isn't possible to set up a disc method. The main time you need this is when revolving around a vertical line #x=a#