let the region r be the area enclosed by the function f(x)=x², the horizontal line y = - 2 and the vertical…

Explanation:

Step1: Find the base of the rectangle

The base of the rectangle at a given $x$ - value is $b(x)=x^{2}-(-2)=x^{2}+2$ (the distance between the curve $y = x^{2}$ and the line $y=-2$).

Step2: Find the height of the rectangle

The height of the rectangle $h(x)=\frac{1}{2}b(x)=\frac{1}{2}(x^{2}+2)$.

Step3: Set up the volume formula

The volume $V$ of the solid with cross - sectional area $A(x)$ from $x = a$ to $x = b$ is given by $V=\int_{a}^{b}A(x)dx$. Here, $A(x)=b(x)\times h(x)=(x^{2}+2)\times\frac{1}{2}(x^{2}+2)=\frac{1}{2}(x^{2}+2)^{2}$, and $a = 0$, $b = 3$.

Step4: Expand the integrand

Expand $\frac{1}{2}(x^{2}+2)^{2}=\frac{1}{2}(x^{4}+4x^{2}+4)=\frac{1}{2}x^{4}+2x^{2}+2$.

Step5: Integrate the function

$\int_{0}^{3}(\frac{1}{2}x^{4}+2x^{2}+2)dx=\left[\frac{1}{2}\times\frac{x^{5}}{5}+2\times\frac{x^{3}}{3}+2x\right]_{0}^{3}$.

Step6: Evaluate the definite integral

First, substitute $x = 3$: $\frac{1}{10}\times3^{5}+\frac{2}{3}\times3^{3}+2\times3=\frac{243}{10}+18 + 6$. $\frac{243}{10}+18+6=\frac{243}{10}+\frac{180}{10}+\frac{60}{10}=\frac{243 + 180+60}{10}=\frac{483}{10}=48.3$.

Answer:

$48.300$