Find the position of the centre of mass of the solid of revolution formed when the region between the curve y = 2e^x, the lines x = 0 and x = 2, and the x axis is rotated through 360° about the x axis.

Question

Find the position of the centre of mass of the solid of revolution formed when the region between the curve [latex]y = 2e^{x}[/latex], the lines x = 0 and x = 2, and the x axis is rotated through 360° about the x axis.

Accepted Answer

The diagrams show the region and the solid obtained

The volume, V, of the solid is given by

V = [latex]\int {πy² dx}[/latex]

= π [latex]\int_{0}^{2}{(2e^{x})^{2} dx}[/latex]

V = 4π[latex]\int_{0}^{2}{e^{2x} dx}[/latex]

= 4π[latex]\left[\frac{e^{2x}}{2} \right]^{2}_{0}[/latex]

= [latex]2π(e^{4} - 1)[/latex].

By symmetry, the y co-ordinate of the centre of mass is 0. The x co-ordinate is given by [latex]\bar{x}[/latex] in

V [latex]\bar{x} = \int{πy^{2}x dx}[/latex]

= π[latex]\int_{0}^{2}{(2e^{x})^{2} x dx}[/latex]

Therefore V [latex]\bar{x} = 4 π \int_{0}^{2}{xe^{2x} dx} . [/latex] ①

This integral requires the technique of integration by parts, expressed by the general formula

[latex]\int_{a}^{b}{u \frac{dv}{dx} dx} = \left[uv \right]^{b}_{a} - \int_{a}^{b}{v \frac{du}{dx} dx}[/latex].

To find [latex]\int{xe^{2x} dx}[/latex]

let u = x ⇒ [latex]\frac{du}{dx} = 1[/latex],

and let [latex]\frac{dv}{dx} = e^{2x}[/latex] ⇒ [latex]v = \frac{e^{2x}}{2}[/latex]

The limits are a = 0 and b = 2.

Substituting these in the general formula, you obtain

[latex]\int_{0}^{2}{xe^{2x} dx} = \left[x \frac{e^{2x}}{2} \right]^{2}_{0} - \int_{0}^{2}{\frac{e^{2x}}{2} × 1 × dx}[/latex]

= [latex]\left[\frac{1}{2} xe^{2x} \right]^{2}_{0} - \left[\frac{1}{4}e^{2x} \right]^{2}_{0}[/latex]

= [latex](e^{4} - 0) - (\frac{1}{4}e^{4} - \frac{1}{4})[/latex]

= [latex]\frac{1}{4} (3e^{4} + 1)[/latex].

Substituting this and [latex]V = 2π(e^{4} - 1)[/latex] in ①

V[latex]\bar{x} = 4 π \int_{0}^{2}{xe^{2x} dx} [/latex]

gives [latex]2π(e^{4} - 1) \bar{x} = 4 π × \frac{1}{4}(3e^{4} + 1)[/latex]

⇒ [latex]\bar{x} = \frac{(3e^{4} + 1)}{2(e^{4} - 1)}[/latex]

= 1.54 (correct to 2 d.p.).

The centre of mass is at (1.54, 0).

Note
If the answer is required in decimal form, it is best to substitute the numerical values of constants such as e and π later rather than sooner. Notice that π cancels in this example.

Question 8.12: Find the position of the centre of mass of the solid of revo...

Related Answered Questions

Verified Answer:

Find the co-ordinates of the centroid of the region between the curve y = 4 – x² and the positive x and y axes. ...

Verified Answer:

Find the position of the centre of mass of a uniform semi-circular lamina of radius r. ...

Verified Answer:

A traffic cone is modelled as a hollow cone of height 45 cm standing on a heavy cylindrical base of radius 20 cm and height 5 cm. The mass of the base is twice that of the cone. i) Find the position of the centre of mass of the traffic cone. ii) What is the angle of the steepest slope it can be ...

Verified Answer:

Find the centre of mass of a solid hemisphere of radius r. ...

Verified Answer:

Find the volume of a spherical ball of radius 2 cm. ...

Verified Answer:

Find the volume of revolution formed when the region between the curve y = x² + 2 , the x axis and the lines x = 1 and x = 3 is rotated completely about the x axis. ...

Verified Answer:

A metal disc of radius 15 cm has a hole of radius 5 cm cut in it as shown in figure 8.15. Find the centre of mass of the disc. ...

Verified Answer:

Verified Answer:

Find the centre of mass of a body consisting of a square plate of mass 3 kg and side length 2 m, with small objects of mass 1 kg, 2 kg, 4 kg and 5 kg at the corners of the square. ...

Verified Answer: