Maths Volume Integration question (1 Viewer)

flowerp · Jun 7, 2019

Hi everyone, I would like help with the following question from Cambridge Yr 11 Ex 12E please:
Find the volume of the rotation of the area in the first quadrant under y = 1 and above y = 2 - 2/x about the x-axis.

Thank you sooooo much!!

Drongoski · Jun 8, 2019

OK - I'll answer this soon.

Drongoski · Jun 8, 2019

The area above the x-axis between y = 1 and y = 2 - 2/x can be split into 2 parts, A and B
The latter curve cuts the x-axis at x=1, and the line y=1 at x=2. Rotating about the x-axis, A generates the volume of a cylinder (height 1, radius 1) = $\pi $ $ u^3$ .
B generates:
$\pi \int ^2 _1 1^2 - (2 - \frac {2}{x})^2 dx = \pi \int ^- _- (-3 + \frac {8}{x} - \frac {4}{x^2}) dx = \pi(-5 + 8ln 2 )\\ \\ \therefore $ total volume$ = \pi + \pi(-5 + 8ln 2) = 4\pi(2ln2 - 1) $ $ u^3$

This is my 1st solution post in half a year, after having trouble logging on previously!

flowerp · Jun 8, 2019

Drongoski said:
The area above the x-axis between y = 1 and y = 2 - 2/x can be split into 2 parts, A and B
The latter curve cuts the x-axis at x=1, and the line y=1 at x=2. Rotating about the x-axis, A generates the volume of a cylinder (height 1, radius 1) = $\pi $ $ u^3$ .
B generates:
$\pi \int ^2 _1 1^2 - (2 - \frac {2}{x})^2 dx = \pi \int ^- _- (-3 + \frac {8}{x} - \frac {4}{x^2}) dx = \pi(-5 + 8ln 2 )\\ \\ \therefore $ total volume$ = \pi + \pi(-5 + 8ln 2) = 4\pi(2ln2 - 1) $ $ u^3$

This is my 1st solution post in half a year, after having trouble logging on previously!

Thank you soooo much for your assistance !!!!! Much appreciated

Maths Volume Integration question (1 Viewer)

flowerp

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Drongoski

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Drongoski

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flowerp

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