Question

In Exercises \(11-22,\) evaluate the iterated integral. $$ \int_{0}^{2} \int_{3 y^{2}-6 y}^{2 y-y^{2}} 3 y d x d y $$

Short answer

The value of the iterated integral is 8.

Step-by-step solution

Evaluate the Inner Integral

First, consider the inner integral which is with respect to x: \(\int_{3 y^{2}-6 y}^{2 y-y^{2}} 3 y d x\). Since the integrand 3y is only a function of y, the integral with respect to x can be evaluated as 3y times the difference of the upper and lower limit of x, which results in: 3y\((2y - y^2) - (3y^2 - 6y)\) = \(3y(-2y^2 + 8y)\).

Evaluate the Outer Integral

Next, consider the outer integral: \(\int_{0}^{2} 3y(-2y^2 + 8y) dy\). This can be computed by regular methods of integration, leading to: \((-(1/2)y^4 + 2y^3)\) evaluated from 0 to 2. Substituting the limits we get: \(-(1/2)(2)^4 + 2(2)^3 - (-(1/2)(0)^4 + 2(0)^3) = -8 + 16 = 8\)

Final Result

After evaluating both the inner and outer integral, the evaluation of the entire iterated integral is 8.