find volume generated.?

Question

given y^2=4x and x^2=4y intersect at (0,0) and (4,4).aRea between curves rotated completely about x axis.find volume generated...answer given is 19.2 pi.can u get the answer?

Answer 1

You are correct.

y² = 4x
x² = 4y

Rewriting

y = 2√x
y = x² / 4

V = volume of rotation

V = ∫π(R² - r²)dx | Evaluated on [0,4]

V = ∫π(R² - r²)dx
= ∫π[(2√x)² - (x²/4)²]dx
= π∫[4x - (x^4)/16)]dx
= π[2x² - (x^5)/80] | Evaluated on [0,4]
= π[2*16 - 1024/80] - 0
= π(32 - 64/5) = π(32 - 12.8) = 19.2π

Answer 2

Let the volume generated by curve- y^2=4x be V1
and Let the volume generated by curve- x^2=4y be V2

then...

V1=Calculus(0 to 4) [ f(x)^2*pi ] dx=
=Calculus(0 to 4) [ y^2*pi] dx
=Calculus(0 to 4) [ 4x*pi]dx
=32pi

V2=Calculus(0 to 4) [ f(x)^2*pi ] dx=
=Calculus(0 to 4) [ x^4/16*pi ]dx
=pi/16*204.8=12.8pi

so V=V1-V2=32pi-12.8pi=19.2pi

Good luck for you....

Answer 3

V = π ∫ 4x dx - π ∫ x^(4) /16 dx--------limits 0 to 4

V = π. [ 4 . x² / 2 ] - (π / 16). [ x^(5) / 5 ]-----limits 0 to 4

V = 32 π - 1024 π / 80

V = 32 π - 12.8 π

V = 19.2 π