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

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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:

.03x^2+7.1x+350

-6.593x+389

389

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 2pi*Integral[(x+220)(.03x^2+7.1x+350)] 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:
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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