Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=x^2, y=1; about y=9
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Since we are rotating about horizontal line, we will use washer method
and integrate with respect to x
The curve y = x² and line y = 1 intersect at point (−1, 1) and (1, 1)
So we integrate from x = −1 to x = 1
On the interval −1 < x < 1, x² < 1
But since axis of rotation y = 9 is higher than both the parabola and line, then:
outer radius: R = 9 − x²
inner radius: r = 9 − 1 = 8
V = π ∫ₐᵇ (R² − r²) dx
V = π ∫₋₁¹ ((9 − x²)² − 8²) dx
V = π ∫₋₁¹ (x⁴ − 18x² + 17) dx
V = π (1/5 x⁵ − 6x³ + 17x) |₋₁¹
V = π (1/5 − 6 + 17) − (−1/5 + 6 − 17)
V = π (56/5 + 56/5)
V = 112π/5
and integrate with respect to x
The curve y = x² and line y = 1 intersect at point (−1, 1) and (1, 1)
So we integrate from x = −1 to x = 1
On the interval −1 < x < 1, x² < 1
But since axis of rotation y = 9 is higher than both the parabola and line, then:
outer radius: R = 9 − x²
inner radius: r = 9 − 1 = 8
V = π ∫ₐᵇ (R² − r²) dx
V = π ∫₋₁¹ ((9 − x²)² − 8²) dx
V = π ∫₋₁¹ (x⁴ − 18x² + 17) dx
V = π (1/5 x⁵ − 6x³ + 17x) |₋₁¹
V = π (1/5 − 6 + 17) − (−1/5 + 6 − 17)
V = π (56/5 + 56/5)
V = 112π/5