Question

Consider the following. y = 2xe−x, y = 0, x = 3; about the y-axis (a) Set up an integral for the volume V of the solid obtained by rotating the region bounded by the given curve about the specified axis. V = 0 dx (b) Use your calculator to evaluate the integral correct to five decimal places.

Answer 1

Using the shell method, the volume would be

$\displaystyle2\pi\int_0^3x(2xe^{-x})\,\mathrm dx=4\pi\int_0^3x^2e^{-x}\,\mathrm dx$

Each shell contributes a volume of approximately 2πrh, where r is the distance from the axis of revolution to the shell (x) and h is the height of the shell given by the vertical distance between y = 2x e^(-x) and y = 0.

If you go on to use your calculator, you'll find the volume is approximately 14.49681.

Answer 2

a) The integral for the volume of the solid of revolution is $V = 4\pi\int\limits^3_0 {x^{2}\cdot e^{-x}} \, dx$.

b) The volume of this solid of revolution is approximately 1.15362 cubic units.

Determination of the volume of a solid of revolution by integration

a) Let be $f(x)$ an integrable function for the interval $[a,b] \in \mathbb{R}$, the volume of  the solid of revolution ($V$), in cubic units, about the y-axis is:

$V = 2\pi\int\limits^b_a {x\cdot f(x)} \, dx$ (1)

Where:

• $a$ - Lower bound
• $b$ - Upper bound
• $f(x)$ - Integrable function

Now we obtain the integral equation for the volume of the solid of revolution: ($a = 0$, $b = 3$, $f(x) = 2\cdot x\cdot e^{-x}$)

$V = 4\pi\int\limits^3_0 {x^{2}\cdot e^{-x}} \, dx$ (2)

The integral for the volume of the solid of revolution is $V = 4\pi\int\limits^3_0 {x^{2}\cdot e^{-x}} \, dx$. $\blacksquare$

b) Now we evaluate the integral by using a math tool (i.e. Wolfram Alpha). According this tool, the volume of this solid of revolution is approximately 1.15362 cubic units. $\blacksquare$

To learn more on solids of revolution, we kindly invite to check this verified question: https://brainly.com/question/338504