Evaluate the integral. \( \int 2 y^{2} \sqrt{9-2 y^{3}} d y \) \( \int 2 y^{2} \sqrt{9-2 y^{3}} d y=\square \)

We start with the integral \[ \int 2y^2 \sqrt{9-2y^3}\,dy. \] **Step 1.** Use the substitution \[ u = 9-2y^3. \] Then, differentiate with respect to \(y\): \[ \frac{du}{dy} = -6y^2 \quad \Longrightarrow \quad du = -6y^2\,dy. \] **Step 2.** Solve for \(y^2\,dy\): \[ y^2\,dy = -\frac{1}{6}\,du. \] Substitute this into the integral. Note that \[ 2y^2\,dy = 2\left(-\frac{1}{6}\,du\right) = -\frac{1}{3}\,du. \] Thus, the integral becomes \[ -\frac{1}{3}\int \sqrt{u}\,du. \] **Step 3.** Evaluate the integral: \[ \int \sqrt{u}\,du = \int u^{\frac{1}{2}}\,du = \frac{u^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3}u^{\frac{3}{2}}. \] So, we have \[ -\frac{1}{3}\int \sqrt{u}\,du = -\frac{1}{3}\cdot\frac{2}{3} u^{\frac{3}{2}} = -\frac{2}{9} u^{\frac{3}{2}}. \] **Step 4.** Substitute back \(u = 9-2y^3\): \[ -\frac{2}{9}(9-2y^3)^{\frac{3}{2}}+C. \] Thus, the evaluated integral is \[ \int 2y^2 \sqrt{9-2y^3}\,dy = -\frac{2}{9}(9-2y^3)^{\frac{3}{2}}+C. \]