Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.
y = 1/x^2, y = 0, x = 2, x = 7
about y = -3.
So I said pi * integral from 2 to 7 of (1/x^2)^2 -3...what am I doing wrong because I know the answer isn't right. Can someone please show me steps and then an answer?
y = 1/x^2, y = 0, x = 2, x = 7
about y = -3.
So I said pi * integral from 2 to 7 of (1/x^2)^2 -3...what am I doing wrong because I know the answer isn't right. Can someone please show me steps and then an answer?
-
Use washer method: π (R² - r²) dx, where
R = distance from upper function to axis of rotation: 1/x² - (-3) = 1/x² + 3
r = distance from lower function (y=0) to axis of rotation(y=-3): 0 - (-3) = 3
V = π ∫₂⁷ ((1/x² + 3)² - (3)²) dx
V = π ∫₂⁷ (1/x⁴ + 6/x² + 9 - 9) dx
V = π ∫₂⁷ (1/x⁴ + 6/x²) dx
V = π ∫₂⁷ (1/x⁴ + 6/x²) dx
V = π (-1/(3x³) - 6/x) |₂⁷
V = π (-1/1029 - 6/7 + 1/24 + 3)
V = π (-8/8232 - 7056/8232 + 343/8232 + 24696/8232)
V = 17975π/8232
V = 6.85983
R = distance from upper function to axis of rotation: 1/x² - (-3) = 1/x² + 3
r = distance from lower function (y=0) to axis of rotation(y=-3): 0 - (-3) = 3
V = π ∫₂⁷ ((1/x² + 3)² - (3)²) dx
V = π ∫₂⁷ (1/x⁴ + 6/x² + 9 - 9) dx
V = π ∫₂⁷ (1/x⁴ + 6/x²) dx
V = π ∫₂⁷ (1/x⁴ + 6/x²) dx
V = π (-1/(3x³) - 6/x) |₂⁷
V = π (-1/1029 - 6/7 + 1/24 + 3)
V = π (-8/8232 - 7056/8232 + 343/8232 + 24696/8232)
V = 17975π/8232
V = 6.85983