integration frustration (1 Viewer)

[darshil]

not really frustrated but sounds cool doesnt it?

I was just looking at this question and coudnt really get it:

7) it is a Region bounded by parabola y=(x-2)^2 and the coordinate axis about the X axis. Find the volume of the solid formed.

Also what are some tips for visualise these revolutions of a curve and its volume?

Thanks for all the help

 

[zazzy1234]

not really frustrated but sounds cool doesnt it?

I was just looking at this question and coudnt really get it:

7) it is a Region bounded by parabola y=(x-2)^2 and the coordinate axis about the X axis. Find the volume of the solid formed.

Also what are some tips for visualise these revolutions of a curve and its volume?

Thanks for all the help

1. first square both sides, so it should be y^2=(x-2)^4

2. then u intergrate using x=2 and x=0

 

[Timothy.Siu]

1. first square both sides, so it should be y^2=(x-2)^4

2. then u intergrate using x=2 and x=0

x pi

 

[zazzy1234]

x pi

where did the "x" come from

 

[gurmies]

where did the "x" come from

multiplied by pi, not the pronumeral x

 

[zazzy1234]

multiplied by pi, not the pronumeral x

yeh i guess ur right

 

[darshil]

I dont think I get it, sorry

 

[zazzy1234]

I dont think I get it, sorry

look its simple. To find such volume u must always square both sides (it's a rule) so it turns out to be

y^2= (x-2)^4

then u

intergrate (x-2)^4 using the x=0 and x=2 and put in a pi at the end of ur answer

 

[darshil]

thanks mate i get it now, cheers!

 

[addikaye03]

Its just:

V=pi int. (b--->a) x^2 dy (y-axis revolution)

OR/ V=pi int(b--->a) y^2 dx ( x-axis revolution)

 

[maths-bbqz]

Hey, the title almost rhymes

 

[Official]

Haha yeh

 

[darshil]

hahhah that was the point (=