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The region bounded by y = 3x - 9 and y = - 3x^2 + 18 x - 9 revolved about the

y - axis. Find the volume of the solid generated.

User Josmith
by
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1 Answer

4 votes
Answer:h(x) =
the volume of the solid generated is approximately 468.8π cubic units.

Explanation:
To find the volume of the solid generated when the region bounded by y = 3x - 9 and y = -3x^2 + 18x - 9 is revolved about the y-axis, we can use the method of cylindrical shells.

First, we need to find the limits of integration. The two curves intersect when:

3x - 9 = -3x^2 + 18x - 9
3x^2 - 15x = 0
x(3x - 15) = 0

So the curves intersect at x = 0 and x = 5.
V = ∫[a,b] 2πr(x)h(x) dx
h(x) = (3x - 9) - (-3x^2 + 18x - 9) = 3x^2 - 15x

Substituting in the limits of integration and evaluating the integral, we get:

V = ∫[0,5] 2πx(3x^2 - 15x) dx
= 2π ∫[0,5] (3x^3 - 15x^2) dx
= 2π [(3/4)x^4 - 5x^3] |[0,5]
= 2π [(3/4)(5^4) - 5(5^3)]
= 2π (375/4)
= 468.8π

Therefore, the volume of the solid generated is approximately 468.8π cubic units.


User Christian Sloper
by
8.0k points