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

let the region r be the area enclosed by the function f(x)=x², the horizontal line y = -2 and the vertical lines x = 0 and x = 3. if the region r is the base of a solid such that each cross - section perpendicular to the x - axis is a rectangle whose height is half the length of its base in the region r, find the volume of the solid. you may use a calculator and round to the nearest thousandth.

let the region r be the area enclosed by the function f(x)=x², the horizontal line y = -2 and the vertical lines x = 0 and x = 3. if the region r is the base of a solid such that each cross - section perpendicular to the x - axis is a rectangle whose height is half the length of its base in the region r, find the volume of the solid. you may use a calculator and round to the nearest thousandth.

Answer

Explanation:

Step1: Find the base of the cross - section

The base of the rectangle cross - section perpendicular to the $x$-axis is given by the difference between the upper and lower functions. Here, the upper function is $y = x^{2}$ and the lower function is $y=-2$. So, the base $b$ of the rectangle at a given $x$ is $b=x^{2}-(-2)=x^{2}+2$.

Step2: Find the height of the cross - section

The height $h$ of the rectangle is half the length of its base. So, $h=\frac{1}{2}(x^{2}+2)$.

Step3: Find the area of the cross - section

The area $A(x)$ of a rectangle is $A(x)=b\times h$. Substituting the values of $b$ and $h$, we get $A(x)=(x^{2}+2)\times\frac{1}{2}(x^{2}+2)=\frac{1}{2}(x^{2}+2)^{2}=\frac{1}{2}(x^{4}+4x^{2}+4)$.

Step4: Calculate the volume using the integral

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, $a = 0$, $b = 3$, and $A(x)=\frac{1}{2}(x^{4}+4x^{2}+4)$. So, $V=\int_{0}^{3}\frac{1}{2}(x^{4}+4x^{2}+4)dx=\frac{1}{2}\int_{0}^{3}(x^{4}+4x^{2}+4)dx$. We know that $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). $\frac{1}{2}\left[\frac{x^{5}}{5}+4\times\frac{x^{3}}{3}+4x\right]_{0}^{3}=\frac{1}{2}\left(\frac{3^{5}}{5}+\frac{4\times3^{3}}{3}+4\times3\right)$. $\frac{1}{2}\left(\frac{243}{5}+36 + 12\right)=\frac{1}{2}\left(\frac{243}{5}+48\right)=\frac{1}{2}\left(\frac{243+240}{5}\right)=\frac{1}{2}\times\frac{483}{5}=\frac{483}{10}=48.300$.

Answer:

$48.300$