Volume of revolution
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Volume of revolution

[From: ] [author: ] [Date: 11-10-22] [Hit: ]
.so its worth going through my work with a fine toothed comb, but thats not my job here, so I probably wont try to find the error.-I cannot stress enough how important it is to draw a picture for these types of problems........

volume of a THIN cylindrical shell (with width dr):

dV = 2πr*h*dr

r is easy, it's still (3 - x). h on the other hand, is a little more difficult. If you draw it out, you can see the the height is the difference in the y values:

h = y₁ - y₀
-->

dr = dx (since it's perpendicular to the axis), so we have:

dV = 2π(3 - x)*(y₁ - y₀)*dx
-->

Now we need to find y₁ - y₀:

y = ±√(x/3)
--> clearly +√(x/3) is y "top" and -√(x/3) is y bottom:

y₁ - y₀ = √(x/3) - -√(x/3) = √(x/3) + √(x/3) = 2√(x/3)
-->

dV = 4π(3 - x)*√(x/3)*dx = 4π/√3 * (3 - x)√x dx
--> distribute to integrate:

(3 - x)√x = 3√x - x^(3/2)
--> integrate

3*(2/3)*x^(3/2) - 2/5 * x^(5/2) = 2x^(3/2) - 2/5 * x^(5/2)
-->

x goes from 0 to x = 3, so that's the integral (just plug in x = 3):

3^(3/2) = 3 * √3
3^(5/2) = 3² * √3 = 9√3
-->

2*3√3 - 2/5 * 9√3 = 2*3(√3 - 3/5 √3) = 6√3 * (2/5) = 12√3/5
--> plug that in with the constants from above:

4π/√3 * 12√3/5 = 4*12π/5 = 48π/5

OK, so I've made a mistake somewhere, either above (in the disk) method, or here, in the shell method (or worse, in both). These SHOULD give the same results...so it's worth going through my work with a fine toothed comb, but that's not my job here, so I probably won't try to find the error.

-
I cannot stress enough how important it is to draw a picture for these types of problems...because if you don't have a picture in your head, then, you don't understand the problem

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distance from outside shell to x=3 is 3y^2 - 3.

the surface is pi * (3y^2 - 3)^2
=> the volume element is (pi * (3y^2 - 3)^2) * dy =>

int(dy pi (3y^2 - 3)^2, y=0..3) =
pi * (9/5y^5-6y^3+9y)|y=0..3 =
I don't have a calculator :)

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No, use the disc method.

V = π ∫ [f(y)]² dy

V = π ∫ [3y² - 3]² dy[-1,1]
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