1. **Problem Statement:** Find the volume of the solid bounded by the planes $E_1: y = x^2$, $E_2: y + z = 16$, and $E_3: z = 0$. 2. **Boundaries:** - From $E_1$, we have $y = x^2$. - From $E_3$, $z = 0$. - From $E_2$, $z = 16 - y$. 3. **Determine the projection in the xy-plane:** Since $z$ ranges from 0 to $16 - y$, and $y$ must satisfy $y \geq x^2$ (from $E_1$) and $y \leq 16$ (since $z \geq 0$ implies $16 - y \geq 0$), the region in the xy-plane is: $$x^2 \leq y \leq 16$$ and $x$ ranges over all real numbers such that $x^2 \leq 16$, i.e., $$-4 \leq x \leq 4$$ 4. **Volume integral setup:** $$\text{Volume} = \int_{x=-4}^{4} \int_{y=x^2}^{16} \int_{z=0}^{16 - y} dz\, dy\, dx$$ 5. **Evaluate the innermost integral:** $$\int_0^{16 - y} dz = 16 - y$$ 6. **Volume integral reduces to:** $$\int_{-4}^4 \int_{x^2}^{16} (16 - y) dy dx$$ 7. **Evaluate the inner integral:** $$\int_{x^2}^{16} (16 - y) dy = \left[16y - \frac{y^2}{2}\right]_{y=x^2}^{16} = \left(16 \times 16 - \frac{16^2}{2}\right) - \left(16 x^2 - \frac{x^4}{2}\right)$$ Calculate constants: $$16 \times 16 = 256$$ $$\frac{16^2}{2} = \frac{256}{2} = 128$$ So, $$256 - 128 - 16 x^2 + \frac{x^4}{2} = 128 - 16 x^2 + \frac{x^4}{2}$$ 8. **Volume integral becomes:** $$\int_{-4}^4 \left(128 - 16 x^2 + \frac{x^4}{2}\right) dx$$ 9. **Evaluate each term separately:** - $\int_{-4}^4 128 dx = 128 \times (4 - (-4)) = 128 \times 8 = 1024$ - $\int_{-4}^4 -16 x^2 dx = -16 \int_{-4}^4 x^2 dx = -16 \times \left[\frac{x^3}{3}\right]_{-4}^4 = -16 \times \left(\frac{64}{3} - \frac{-64}{3}\right) = -16 \times \frac{128}{3} = -\frac{2048}{3}$ - $\int_{-4}^4 \frac{x^4}{2} dx = \frac{1}{2} \int_{-4}^4 x^4 dx = \frac{1}{2} \times \left[\frac{x^5}{5}\right]_{-4}^4 = \frac{1}{2} \times \left(\frac{1024}{5} - \frac{-1024}{5}\right) = \frac{1}{2} \times \frac{2048}{5} = \frac{1024}{5}$ 10. **Sum all parts:** $$1024 - \frac{2048}{3} + \frac{1024}{5} = \frac{1024 \times 15}{15} - \frac{2048 \times 5}{15} + \frac{1024 \times 3}{15} = \frac{15360 - 10240 + 3072}{15} = \frac{6144}{15} = \frac{409.6}{1}$$ So, $$\text{Volume} = \frac{6144}{15}$$ **Final answer:** $$\boxed{\text{Volume} = \frac{6144}{15}}$$