How Do You Calculate the Volume of an Arch Dam Using the Shell Method?

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SUMMARY

The discussion focuses on calculating the volume of an arch dam using the shell method, specifically for a rotation of 150 degrees around an axis of 150. The user plans to split the cross-section into two triangles and one rectangle, applying the shell method to each section. The integral used is 2π∫[(x+220)(0.03x² + 7.1x + 350)] from -70 to -16, adjusted for the 150-degree rotation by multiplying by (150/360). The approach is confirmed to be on the right track for calculating the volume.

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Homework Statement



The problem is number 4 at this link: http://college.cengage.com/mathematics/larson/calculus_early/2e/students/downloads/mws6a.pdf

The axis of revolution is 150 and the rotation is 150 degrees.

Homework Equations



Using the shell method.

My offset is x+220.

The equations that make up the cross section:

[tex].03x^2+7.1x+350[/tex]

[tex]-6.593x+389[/tex]

[tex]389[/tex]

The Attempt at a Solution



I planned on splitting the cross section into three parts, which are the 2 triangles and rectangles, and apply the shell method on each of these. So first I did [tex]2pi*Integral[(x+220)(.03x^2+7.1x+350)][/tex] with the bounds being -70 to -16. Then I multiplied this answer by (150/360) since it's only 150 degree rotation. In each of my applications of the shell method, the p(x), or the distance to the axis of revolution stayed at x+220. Am I on the right track if I do this for each of the 3 sections?
 
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Welcome to PF!

Hi Sentral! Welcome to PF! :smile:

(have a pi: π and an integral: ∫ and try using the X2 tag just above the Reply box :wink:)

I'm not sure what you're doing (what's 220? :confused:) …

the shell method involves finding the volume of the arc-shaped slice (it will be 150/360 of the volume of a complete cylinder) from x to x + dx …

it will be something times dx …

and then integrating. :smile:
 

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