a volume is described as follows: 1. the base is the region bounded by (y = e^{2.7x}), (y = 2.7x^{2}+0.9)…

a volume is described as follows: 1. the base is the region bounded by (y = e^{2.7x}), (y = 2.7x^{2}+0.9) and (x = 1); 2. every cross - section perpendicular to the x - axis is a square. find the volume of this object. volume =

a volume is described as follows: 1. the base is the region bounded by (y = e^{2.7x}), (y = 2.7x^{2}+0.9) and (x = 1); 2. every cross - section perpendicular to the x - axis is a square. find the volume of this object. volume =

Answer

Explanation:

Step1: Determine the side - length of the square cross - section

The side - length $s$ of each square cross - section perpendicular to the $x$ - axis is given by the difference between the two functions. So, $s = e^{2.7x}-(2.7x^{2}+0.9)$.

Step2: Find the area of the square cross - section

The area $A(x)$ of a square is $A(x)=s^{2}$. So, $A(x)=(e^{2.7x}-2.7x^{2}-0.9)^{2}=e^{5.4x}-5.4x^{2}e^{2.7x}-1.8e^{2.7x}+7.29x^{4}+4.86x^{2}+0.81$.

Step3: Calculate the volume using the integral

The volume $V$ of the solid with cross - sectional area $A(x)$ from $x = 0$ (the left - hand intersection point of $y = e^{2.7x}$ and $y = 2.7x^{2}+0.9$ which is approximately $x = 0$) to $x = 1$ is given by the integral $V=\int_{0}^{1}A(x)dx=\int_{0}^{1}(e^{5.4x}-5.4x^{2}e^{2.7x}-1.8e^{2.7x}+7.29x^{4}+4.86x^{2}+0.81)dx$. We use integration by parts for $\int x^{2}e^{2.7x}dx$. Let $u = x^{2}$, $dv=e^{2.7x}dx$, then $du = 2xdx$, $v=\frac{1}{2.7}e^{2.7x}$. $\int x^{2}e^{2.7x}dx=\frac{x^{2}}{2.7}e^{2.7x}-\frac{2}{2.7}\int xe^{2.7x}dx$. For $\int xe^{2.7x}dx$, let $u = x$, $dv=e^{2.7x}dx$, then $du=dx$, $v = \frac{1}{2.7}e^{2.7x}$, and $\int xe^{2.7x}dx=\frac{x}{2.7}e^{2.7x}-\frac{1}{2.7^{2}}e^{2.7x}+C$. $\int e^{5.4x}dx=\frac{1}{5.4}e^{5.4x}+C$, $\int e^{2.7x}dx=\frac{1}{2.7}e^{2.7x}+C$, $\int x^{4}dx=\frac{1}{5}x^{5}+C$, $\int x^{2}dx=\frac{1}{3}x^{3}+C$, $\int 0.81dx = 0.81x+C$. $V=\left[\frac{1}{5.4}e^{5.4x}-5.4\left(\frac{x^{2}}{2.7}e^{2.7x}-\frac{2}{2.7}\left(\frac{x}{2.7}e^{2.7x}-\frac{1}{2.7^{2}}e^{2.7x}\right)\right)-1.8\times\frac{1}{2.7}e^{2.7x}+7.29\times\frac{1}{5}x^{5}+4.86\times\frac{1}{3}x^{3}+0.81x\right]_{0}^{1}$ $V=\frac{1}{5.4}e^{5.4}- 2\left(e^{2.7}-\frac{2}{2.7}\left(\frac{1}{2.7}e^{2.7}-\frac{1}{2.7^{2}}e^{2.7}\right)\right)-\frac{2}{3}e^{2.7}+\frac{7.29}{5}+1.62 + 0.81-\left(\frac{1}{5.4}- \frac{2}{2.7^{3}}e^{0}-\frac{2}{3}e^{0}+0 + 0+0\right)$ After calculating: $V\approx119.07$

Answer:

$119.07$