Rotating-a-curve-y-x-about-the-x-axis-produces-a-head-light-as-shown-below-What-is-the-area-of-disc-at-any-x-

Question Number 15036 by Tinkutara last updated on 07/Jun/17

Rotating a curve y = \sqrt{x} about the x -

axis produces a “ head {light}^{″} as shown

below .

What is the area of disc at any x ?

Commented by Tinkutara last updated on 07/Jun/17

Commented by Tinkutara last updated on 07/Jun/17

In the above problem what is the

volume of this head light where x varies

from 0 to 2 ?

Answered by ajfour last updated on 07/Jun/17

A r e a o f d i s c = π {(\sqrt{x})}^{2}

V o l u m e = \int_{0}^{2} (π x) d x

= (\frac{π x^{2}}{2}) ∣_{x = 0}^{x = 2} = \frac{π}{2} {(\sqrt{2})}^{2} (2) = 2 π .

Commented by Tinkutara last updated on 07/Jun/17

But will the disc be circular so that we

can apply area = π r^{2} or will they be

elliptical ?

Commented by ajfour last updated on 07/Jun/17

a n e l e m e n t a r y l a y e r o f d i s c o f

a t l o c a t i o n t h a s i t s t h i c k n e s s d t

a n d r a d i u s \sqrt{t} . I t s v o l u m e w i l l b e

d V = (A r e a \times t h i c k n e s s)

= π {(\sqrt{t})}^{2} d t

t o o b t a i n t h e s u m o f v o l u m e s

o f s u c h c o n s e c u t i v e l a y e r s s t a r t i n g

a t t = 0 t i l l t = x, w e i n t e g r a t e t h e

e x p r e s s i o n o f d V w i t h l i m i t s

t = 0, t o t = x .

V (x) = π (\frac{t^{2}}{2}) ∣_{t = 0}^{t = x} = \frac{π x^{2}}{2} .

Commented by ajfour last updated on 07/Jun/17

w h e n y o u r o t a t e a x e s y o u d o n t

s t r e t c h i t, r a d i u s w i l l b e \sqrt{x}, i t

w i l l b e c i c u l a r d i s c .

Commented by Tinkutara last updated on 07/Jun/17

Thanks Sir!