Key Concepts and Formulas

• Area between curves: The area between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x=b$, where $f(x) \geq g(x)$ on $[a, b]$, is given by $\int_a^b [f(x) - g(x)] dx$.
• Intersection of curves: To find where two curves intersect, set their equations equal to each other and solve for $x$.
• Definite Integral: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$.

Step-by-Step Solution

Step 1: Find the intersection points of the curves $y=3x$ and $2y=27-3x$.

We need to find the $x$ value where the lines intersect. Substitute $y=3x$ into $2y=27-3x$: $2(3x) = 27 - 3x$ $6x = 27 - 3x$ $9x = 27$ $x = 3$ When $x=3$, $y=3(3)=9$. So, the intersection point is $(3, 9)$.

Step 2: Find the intersection points of the curves $y=3x$ and $y=3x - x\sqrt{x}$.

We need to find the $x$ value where the curves intersect. Set the equations equal: $3x = 3x - x\sqrt{x}$ $0 = -x\sqrt{x}$ $x\sqrt{x} = 0$ $x = 0$ When $x=0$, $y=3(0)=0$. So, the intersection point is $(0, 0)$.

Step 3: Find the intersection points of the curves $2y=27-3x$ and $y=3x - x\sqrt{x}$.

Substitute $y=3x - x\sqrt{x}$ into $2y=27-3x$: $2(3x - x\sqrt{x}) = 27 - 3x$ $6x - 2x\sqrt{x} = 27 - 3x$ $9x - 2x\sqrt{x} = 27$ $9x - 2x^{3/2} = 27$ Let $x = t^2$, then the equation becomes: $9t^2 - 2t^3 = 27$ $2t^3 - 9t^2 + 27 = 0$ By observation, $t=3$ is a root, since $2(3)^3 - 9(3)^2 + 27 = 54 - 81 + 27 = 0$. Since $x = t^2$, we have $x = 3^2 = 9$. When $x=9$, $y = 3(9) - 9\sqrt{9} = 27 - 9(3) = 27 - 27 = 0$. So the intersection point is $(9, 0)$.

Step 4: Set up the integrals for the area.

The region is bounded by the three curves. The area can be found by integrating from $x=0$ to $x=3$ and from $x=3$ to $x=9$.

From $x=0$ to $x=3$, $3x$ is above $3x-x\sqrt{x}$. So the area is $\int_0^3 (3x - (3x - x\sqrt{x})) dx = \int_0^3 x\sqrt{x} dx = \int_0^3 x^{3/2} dx$.

From $x=3$ to $x=9$, $\frac{27-3x}{2}$ is above $3x-x\sqrt{x}$. So the area is $\int_3^9 (\frac{27-3x}{2} - (3x - x\sqrt{x})) dx = \int_3^9 (\frac{27}{2} - \frac{3x}{2} - 3x + x^{3/2}) dx = \int_3^9 (\frac{27}{2} - \frac{9x}{2} + x^{3/2}) dx$.

Thus, the total area $A$ is: $A = \int_0^3 x^{3/2} dx + \int_3^9 \left(\frac{27}{2} - \frac{9x}{2} + x^{3/2}\right) dx$

Step 5: Evaluate the integrals.

$A = \left[\frac{2}{5}x^{5/2}\right]_0^3 + \left[\frac{27}{2}x - \frac{9}{4}x^2 + \frac{2}{5}x^{5/2}\right]_3^9$ $A = \frac{2}{5}(3^{5/2}) - 0 + \left(\frac{27}{2}(9) - \frac{9}{4}(9^2) + \frac{2}{5}(9^{5/2})\right) - \left(\frac{27}{2}(3) - \frac{9}{4}(3^2) + \frac{2}{5}(3^{5/2})\right)$ $A = \frac{2}{5}(9\sqrt{3}) + \left(\frac{243}{2} - \frac{729}{4} + \frac{2}{5}(243)\right) - \left(\frac{81}{2} - \frac{81}{4} + \frac{2}{5}(9\sqrt{3})\right)$ $A = \frac{18\sqrt{3}}{5} + \frac{486}{4} - \frac{729}{4} + \frac{486}{5} - \frac{162}{4} + \frac{81}{4} - \frac{18\sqrt{3}}{5}$ $A = \frac{486}{5} + \frac{486 - 729 - 162 + 81}{4} = \frac{486}{5} + \frac{-324}{4} = \frac{486}{5} - 81 = \frac{486 - 405}{5} = \frac{81}{5}$ $A = \frac{81}{5} = 16.2$

Step 6: Calculate 10A.

$10A = 10 \times \frac{81}{5} = 2 \times 81 = 162$

Common Mistakes & Tips

• Incorrect Order of Integration: Be careful about which function is "above" the other. Drawing a diagram is crucial.
• Sign Errors: Pay close attention to signs when subtracting functions and evaluating definite integrals.
• Simplification: Simplifying the expression inside the integral can significantly reduce the complexity of the calculations.

Summary

We first found the intersection points of the given curves. These intersection points defined the limits of integration. Then, we set up the definite integrals to calculate the area of the enclosed region. Finally, we evaluated the integrals to find the area $A$ and computed $10A$, which is 162.

The final answer is \boxed{162}, which corresponds to option (C).