Find the volume of the solid formed by rotating the

Question

Find the volume of the solid formed by rotating the region in the 1st quadrant enclosed by $$y = \frac{1}{11} x\ \ \ \text{and} \ \ \ y = x^{1/3}$$ about the y-axis.

Answer 1

For the sake of closure on the result, I'm working out the numerical values by both the "disk method", which Semsem set up, and by the "shell method".

The "bounding functions" can be expressed as $ \ x = 11y \ $ and $ \ x = y^3 \ $ for applying "disks" (integration in the $ y-$ direction). The intersection points of the curves are given by

$$ \frac{1}{11}x \ - \ x^{1/3} \ = \ 0 \ \ \Rightarrow \ \ x^{1/3} \ (\frac{1}{11}x^{2/3} \ - \ 1 ) \ = \ 0 \ , $$

so we have either $ \ x = 0 \ \Rightarrow \ y = 0 \ \ $ or $ \ \ x^{2/3} = 11 \ \Rightarrow \ x = 11^{3/2} \ \Rightarrow \ y = 11^{1/2} \ . $ The volume integral Semsem shows is then evaluated as

$$ \pi \ \left[ \ \frac{11^2}{3} y^3 \ - \ \frac{1}{7}y^7 \ \right]_0^{11^{1/2}} \ = \ \pi \ \left[ \ \frac{11^2}{3} \cdot (11^{1/2})^3 \ - \ \frac{1}{7} \cdot (11^{1/2})^7 \ - \ 0\ \right] $$

$$ = \ \pi \ \left[ \ \frac{11^2}{3} \cdot (11^{3/2}) \ - \ \frac{1}{7} \cdot (11^{7/2}) \right] \ = \ \pi \ (11^{7/2})\ \left( \frac{1}{3} \ - \ \frac{1}{7} \right) \ = \ \frac{4 \pi}{21} \cdot (11^{7/2}) \ , $$

which may also be written as $ \ \frac{4 \ \cdot \ 11^3 \ \cdot \ \pi}{21} \cdot (11^{1/2}) \ $ or $ \ \frac{5324 \ \pi}{21} \sqrt{11} \ . $

Applying the "shell method" instead, which involves an integration in the $ x-$ direction, the "radius arm" extending from the axis of rotation, the $ \ y-$ axis, is $ \ r = x \ , $ and the height of the shells is given by $ \ x^{1/3} - \frac{1}{11}x \ . $ Here, the volume integral is

$$ 2 \pi \ \int_0^{11^{3/2}} \ x \ (x^{1/3} \ - \ \frac{1}{11}x) \ \ dx \ = \ 2 \pi \ \int_0^{11^{3/2}} \ x^{4/3} \ - \ \frac{1}{11}x^2 \ \ dx $$

$$ = \ 2 \pi \ \left[ \ \frac{3}{7} y^{7/3} \ - \ \frac{1}{33}y^3 \ \right]_0^{11^{1/2}} \ = \ 2 \pi \ \left[ \ \frac{3}{7} \cdot (11^{3/2})^{7/3} \ - \ \frac{1}{33} \cdot (11^{3/2})^3 \ - \ 0\ \right] $$

$$ = \ 2 \pi \ \left[ \ \frac{3}{7} \cdot (11^{7/2}) \ - \ \frac{1}{3 \cdot 11} \cdot (11^{9/2}) \right] \ = \ 2 \pi \ (11^{7/2})\ \left( \frac{3}{7} \ - \ \frac{1}{3} \right) \ = \ \frac{4 \pi}{21} \cdot (11^{7/2}) $$

once again; the decimal expression is indeed $ \ \approx \ 2641.587 \ . $ So the result Andy shows in his comment looks to be correct; it is unclear from what is said here why he believed it to be otherwise. (Perhaps this was a case of a textbook answer written in a different but equivalent form, or a computer-based system rejecting the way the answer was typed...)

Answer 2

$$V=\pi\int_0^\sqrt{11}(11y)^2-(y^3)^2dy$$