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# Find the volume of the solid (Cal2)?

Generated by revolving y=4-x^2 and y =0 about the line y=5.

I am getting conflicting answers.

My integral is ending up to be 2pi (56 - 18x^2 +x^4) dx , from x=0 to x = 2

### 2 Answers

- PopeLv 79 months ago
I like it when they give these with second-degree curves, but usually I do them without the calculus. The cross-section is a parabola segment having these features:

central diameter: x = 0

vertex: V(0, 4)

endpoints of chord: A(-2, 0), B(2, 0)

midpoint of chord: O(0, 0)

area(∆AVB) = 8

area(segment AVB) = (4/3)(8) = 32/3

The centroid of the segment divides central diameter OV in ratio 2:3.

centroid: C(0, 8/5)

radius of path of C = 5 - 8/5 = 17/5

length of path of C = 2π(17/5) = 34π/5

volume of revolution

= (cross-section area)(length of centroid path)

= (32/3)(34π/5)

= 1088π/15

If you are doing this for a calculus class, then I suppose you want to integrate. I do not understand the integrand you used. Here is how it should look:

2π ∫[0,2] {(5 - 0)² - [5 - (4 - x²)]²} dx

= 2π ∫[0,2] (24 - 2x² - x⁴) dx

- rotchmLv 79 months ago
Show ur reasoning & steps that lead to your conflicting answers, and we will show u where u erred.