Find-the-volume-of-the-solid-of-revolution-generated-by-rotating-the-area-bounded-by-y-x-2-x-and-y-x-about-the-y-axis-

Question Number 214509 by depressiveshrek last updated on 10/Dec/24

Find the volume of the solid of revolution

generated by rotating the area bounded

by y = x (2 - x) and y = x about the y - axis .

Commented by mr W last updated on 11/Dec/24

y o u s e e m n e v e r t o g i v e a n y f e e d b a c k

w h e n p e o p l e h a v e a n s w e r e d y o u r

q u e s t i o n s ?

Commented by depressiveshrek last updated on 11/Dec/24

Not true, sometimes I ask for more details

or disagree with the solution / proof . But

why does that matter ?

Commented by mr W last updated on 11/Dec/24

a c t u a l l y i t {d o e s n}^{'} t m a t t e r . i j u s t

t h o u g h t, n o r m a l l y w h e n s o m e b o d y

h a s r e c i e v e d t h e h e l p h e h a d w i s h e d,

{h e}^{'} l l s h o w s o m e r e a c t i o n, n o r m a l l y .

Commented by depressiveshrek last updated on 11/Dec/24

Okay then I will start to give feedback

from now on more frequently .

Answered by mr W last updated on 11/Dec/24

x = y

x^{2} - 2 x + 1 = 1 - y \Rightarrow x = 1 - \sqrt{1 - y}

1 - \sqrt{1 - y} = y \Rightarrow y = 0, 1

V = \int_{0}^{1} π [y^{2} - {(1 - \sqrt{1 - y})}^{2}] d y

= \int_{0}^{1} π (y^{2} + y - 2 + 2 \sqrt{1 - y}) d y

= π {[\frac{y^{3}}{3} + \frac{y^{2}}{2} - 2 y - \frac{4}{3} {(1 - y)}^{\frac{3}{2}}]}_{0}^{1}

= π (\frac{1}{3} + \frac{1}{2} - 2 + \frac{4}{3})

= \frac{π}{6} ✓

Commented by depressiveshrek last updated on 11/Dec/24

Thank you

Commented by mr W last updated on 11/Dec/24

a l t e r n a t i v e m e t h o d w i t h o u t

i n t e g r a l c a l c u l a t i o n :

V_{1} = \frac{2 \times 1 \times 1}{3} \times 2 π \times (\frac{5 \times 1}{8}) = \frac{5 π}{6}

V_{2} = \frac{1 \times 1}{2} \times 2 π \times (\frac{2 \times 1}{3}) = \frac{2 π}{3}

V = V_{1} - V_{2} = \frac{π}{6}

Commented by mr W last updated on 11/Dec/24

Commented by mr W last updated on 11/Dec/24

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