The given function is \( f(x) = -3x^2 + 9x + 54 \). The boundaries are \( x = 0 \) and \( y = 0 \). We need to find the volume of the solid of revolution when this region is rotated about the y-axis.
To use the method of cylindrical shells, we need to express \( x \) as a function of \( y \). Start by solving \( y = -3x^2 + 9x + 54 \) for \( x \).
\[ y = -3x^2 + 9x + 54 \]
\[ 3x^2 - 9x + (54 - y) = 0 \]
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = 3, \, b = -9, \, c = 54 - y \]
\[ x = \frac{9 \pm \sqrt{81 - 12(54 - y)}}{6} \]
\[ x = \frac{9 \pm \sqrt{81 - 648 + 12y}}{6} \]
\[ x = \frac{9 \pm \sqrt{12y - 567}}{6} \]
Find the y-values where the function intersects the y-axis. Set \( x = 0 \):
\[ y = -3(0)^2 + 9(0) + 54 \]
\[ y = 54 \]
So, the limits of integration for \( y \) are from 0 to 54.
Using the method of cylindrical shells, the volume \( V \) is given by:
\[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \]
Since we are rotating around the y-axis, we need to express \( x \) in terms of \( y \):
\[ V = 2\pi \int_{0}^{54} x(y) \, dy \]
From the quadratic formula, we use the positive root:
\[ x = \frac{9 + \sqrt{12y - 567}}{6} \]
Substitute \( x(y) \) into the integral:
\[ V = 2\pi \int_{0}^{54} \left( \frac{9 + \sqrt{12y - 567}}{6} \right) \, dy \]
Simplify the integral:
\[ V = \frac{2\pi}{6} \int_{0}^{54} \left( 9 + \sqrt{12y - 567} \right) \, dy \]
\[ V = \frac{\pi}{3} \int_{0}^{54} \left( 9 + \sqrt{12y - 567} \right) \, dy \]
Separate the integral:
\[ V = \frac{\pi}{3} \left( \int_{0}^{54} 9 \, dy + \int_{0}^{54} \sqrt{12y - 567} \, dy \right) \]
Evaluate each part:
\[ \int_{0}^{54} 9 \, dy = 9y \Big|_{0}^{54} = 9(54) - 9(0) = 486 \]
For the second part, use substitution \( u = 12y - 567 \):
\[ du = 12 \, dy \]
\[ dy = \frac{du}{12} \]
Change the limits of integration:
When \( y = 0 \), \( u = 12(0) - 567 = -567 \)
When \( y = 54 \), \( u = 12(54) - 567 = 81 \)
\[ \int_{0}^{54} \sqrt{12y - 567} \, dy = \int_{-567}^{81} \sqrt{u} \cdot \frac{du}{12} \]
\[ = \frac{1}{12} \int_{-567}^{81} u^{1/2} \, du \]
\[ = \frac{1}{12} \cdot \frac{2}{3} u^{3/2} \Big|_{-567}^{81} \]
\[ = \frac{1}{18} \left( u^{3/2} \Big|_{-567}^{81}
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