The region bounded by f(x)=-1x^2+4x+12, x=0, and y=0 is rotated about the y-axis. Find the volume of the solid of revolution.

Answer 1

-x² + 4x + 12 = 0; x = 6.

Volume v = 2π∫₀⁶x(- x² + 4x + 12)dx = 2π(- x⁴/4 + 4x³/3 + 6x²)₀⁶ = 360π

Answer 2

To find the volume of the solid of revolution formed by rotating the region bounded by the curve \(f(x) = -x^2 + 4x + 12\), the y-axis, x = 0, and y = 0 about the x-axis, you can use the disk method.

The formula for the volume of a solid of revolution using the disk method is given by:

\[V = \pi \int_a^b [f(x)]^2 \, dx\]

where a is 0 and b is the positive x-coordinate where f(x) intersects y=0:

So, first, solve for x in the quadratic equation \(0 = -x^2 + 4x + 12\):

\[ -x^2 + 4x + 12 = 0\]

Factoring the quadratic, we get:

\[-(x - 6)(x + 2) = 0\]

This gives us two roots: \(x = 6\) and \(x = -2\). Therefore, b = 6

Now, use the disk method formula:

\[V = \pi \int_{0}^{6} [f(x)]^2 \, dx\]

\[V = \pi \int_{0}^{6} (-x^2 + 4x + 12)^2 \, dx\]

To evaluate this integral, we need to simplify the expression \((-x^2 + 4x + 12)^2\). Let's denote it as \(g(x)\):

\[g(x) = (-x^2 + 4x + 12)^2\]

Expanding \(g(x)\), we get:

\[g(x) = x^4 - 8x^3 + 16x^2 - 64x + 144\]

Now, we integrate this expression over the given limits:

\[V = \pi \int_{0}^{6} (x^4 - 8x^3 + 16x^2 - 64x + 144) \, dx\]

The result of this integral will be a numerical value. Let me calculate it for you: 3,076.25 cubic units