Setting Up problem using shell method

[Undeadpaladin]

I'm not totally sure how to set up this intergral using the shell method. I don't have to solve for it(That's the easy part, go figure!). My problem is typically when revolving the object about a line not the axis i have some trouble.

The problem is as follows :

y = x^3
y = 1
x = 2
use the shell method about the line y = 10

My fire time posting i'm not sure exactly how to use symbols, but i came up with this

2Pi * Intergral bottom limit = 2, top limit = 8

[ (10-y)(2-y^(1/3) ] dy

The limits for intergration should be right i think, it's the second part i can't seem to figure out exactly. Any help would be welcome!

 

[Deleted member 4993]

I'm not totally sure how to set up this intergral using the shell method. I don't have to solve for it(That's the easy part, go figure!). My problem is typically when revolving the object about a line not the axis i have some trouble.

The problem is as follows :

y = x^3
y = 1
x = 2
use the shell method about the line y = 10

My fire time posting i'm not sure exactly how to use symbols, but i came up with this

2Pi * Intergral bottom limit = 1, top limit = 8

[ (10-y)(2-y^(1/3)) ] dy

The limits for intergration should be right i think, it's the second part i can't seem to figure out exactly. Any help would be welcome!

As far as I can see - only problem is that the lower limit of integration should be (y = ) 1.

 

[soroban]

Hello,Undeadpaladin!

$\displaystyle \text{The region bounded by: }\;y \,= \,x^3,\;y\, =\, 1,\;x \,=\, 2\text{ is revolved about }y \,=\,10.$

$\displaystyle \text{Use the Shell method to find the volume.}$

$\displaystyle \text{I came up with: }\;V \;=\;2\pi\int^8_2(10-y)(2 - y^{\frac{1}{3}})\,dy$

I agree with Subhotosh . . . the lower limit is 1.

          |
       10 + - - - - - - - - - -
          |
        8 + . . . . . . . *
          |               |
          |             *:|
          |           *:::|
        1 + . . . . *-----|
          |         |     |
      ----+---------+-----+------
          |         1     2

Other than that, your set-up is excellent!

You determined the "radius" and "height" correctly . . . Good work!

 

[galactus]

Just for kicks, here's the washer method:

$\displaystyle {\pi}\int_{1}^{2}[9^{2}-(10-x^{3})^{2}]dx$