Solve for n. 10n = 6 Simplify your answer as much as possible.
2 answers:
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Assuming the area below the line y=0 (i.e. x>1) does NOT count, the area to be rotated is shown in the graph attached.
A. Again, using Pappus's theorem,
Area, A = (2/3)*1*(1-(-1))=4/3 (2/3 of the enclosing rectangle, or you can integrate)
Distance of centroid from axis of rotation, R = (2-0) = 2
Volume = 2 π RA = 2 π 2 * 4/3 = 16 π / 3 (approximately = 16.76 units)
B. By integration, using the washer method
Volume =
![=2\pi[x^4/4-2x^3/3-x^2/2+2x]_{-1}^{1}](https://tex.z-dn.net/?f=%3D2%5Cpi%5Bx%5E4%2F4-2x%5E3%2F3-x%5E2%2F2%2B2x%5D_%7B-1%7D%5E%7B1%7D)
![=2\pi([1/4-2/3-1/2+2]-[1/4+2/3-1/2-2])](https://tex.z-dn.net/?f=%3D2%5Cpi%28%5B1%2F4-2%2F3-1%2F2%2B2%5D-%5B1%2F4%2B2%2F3-1%2F2-2%5D%29)
= 16 π /3 as before
A. Again, using Pappus's theorem,
Area, A = (2/3)*1*(1-(-1))=4/3 (2/3 of the enclosing rectangle, or you can integrate)
Distance of centroid from axis of rotation, R = (2-0) = 2
Volume = 2 π RA = 2 π 2 * 4/3 = 16 π / 3 (approximately = 16.76 units)
B. By integration, using the washer method
Volume =
= 16 π /3 as before