Question

The region bounded by f(x)= -2x^2+12x+54, x=0, and y=0 is rotated about the y-axis. Find the volume of the solid of revolution.

Answer 1

3645π cubic units

Explanation:

To find the volume of the solid of revolution created by rotating the region bounded by the curve f(x) = -2x² + 12x + 54, x = 0, and y = 0 about the y-axis, we can use the method of cylindrical shells.

The volume of a solid of revolution using cylindrical shells is given by the integral:

`\displaystyle V=2\pi \int_(a)^(b)\left(xf(x)\right)\; \text{d}x`

As the region is bounded by the x-axis and y-axis, we need to find the x-value where f(x) intersects the x-axis. To do this, set f(x) = 0 and solve for x:

`\begin{aligned}-2x^2 + 12x + 54&=0\\2x^2 - 12x - 54&=0\\x^2 - 6x - 27&=0\\x^2 - 9x+3x - 27&=0\\x(x-9)+3(x - 9)&=0\\(x+3)(x-9)&=0\\\\x+3&=0 \implies x=-3\\x-9&=0 \implies x=9\end{aligned}`

As the region is bounded by the x-axis and y-axis, the valid value of b is x = 9.

Therefore:

`\displaystyle V=2\pi \int_(0)^(9)\left(x\left(-2x^2 + 12x + 54\right)\right)\; \text{d}x`

`\displaystyle V=2\pi \int_(0)^(9)\left(-2x^3 + 12x^2 + 54x\right)\; \text{d}x`

Integrate using the power rule: Increase the power by 1, then divide by the new power.

`\displaystyle V=2\pi \int_(0)^(9)\left(-2x^3 + 12x^2 + 54x\right)\; \text{d}x`

`V=2\pi \left[(-2x^(3+1))/(3+1)+(12x^(2+1))/(2+1)+(54x^(1+1))/(1+1)\right]_(0)^(9)`

`V=2\pi \left[-(1)/(2)x^4+4x^3+27x^2\right]_(0)^(9)`

Evaluate:

`V=2\pi \left[\left(-(1)/(2)(9)^4+4(9)^3+27(9)^2\right)-\left(-(1)/(2)(0)^4+4(0)^3+27(0)^2\right)\right]`

`V=2\pi \left[(3645)/(2)-0\right]`

`V=3645\pi\; \sf cubic\;units`

Therefore, the volume of the solid of revolution is 3645π cubic units.