volumes of solids (1 Viewer)

[.ben]

19) The region bounded by the curve x^(1/2)+y^(1/2)=2 and the co-ordinate axes is rotated about the X-axis. Find the volume of the solid generated.

I got -64pi units cubed but answer is 64/15 x pi units cubed. btw the expansion is a real bugger.

any help would be appreciated.

thanks

 

[Antwan23q]

i got 64/15pi units cubed
are u sure ur expansion is right?

16-32y^(1/2)+24y-8y^(3/2)+y^2 is mine

 

[Riviet]

How do you expand it?

 

[Antwan23q]

How do you expand it?

x^(1/2) + y^(1/2) = 2

x^(1/2)= 2 -y^(1/2)

x=[2-y^(1/2)]^2

V = pi Integrate (0-4) x^2 dy
V=pi Integration (0-4) [2-y^(1/2)]^4 dy

[2-y^(1/2)]^4 using pascals
=16-32y^(1/2)+24y-8y^(3/2)+y^2

V = pi integrate (0-4) [16-32y^(1/2)+24y-8y^(3/2)+y^2] dy
and im to tired to do the rest

 

[Antwan23q]

oh and if u dont know pacals triangle

___1
__121
_ 1331 (to get the next numbers, u add the 2 before, so 1, (1+2)=3, (2+1) = 3, 1)
_14641 (1, (1+3)=4, (3+3)=6, (3+1)=4, 1)
and so on

 

[.ben]

oki my expansion was wrong

thanks!

 

[evette13]

if its rotated about the x axis doesnt it mean the integration is y^2 dx?? i always thought that was the case??