Intergration+Volume Questions (1 Viewer)

[Avenger6]

Hi, came across these today and struggled abit with the following questions:

1) Find the volume of the solid of revolution that is formed when the curve y=x^2+2 is rotated about the x-axis from x=0 to x=2. (The x^2 seems to confuse me here)

and...

2) Find the volume of the solid formed whe the line x+3y-1=0 is rotated about the x-axis from x=0 to x=8.

Help with these questions is greatly appreciated .

 

[Zephyrio]

1) Don't let the x^2 fool you.

The volume is pi multiplied by the integral of y^2 from x = 2 to x = 0.

Which equals pi multiplied by the integral of (x^4 + 4x^2+4) from x = 2 to x = 0.

Then integrate, substitute, extract answer.

2) Make y the subject, then square it.

 

[tommykins]

V = pi int [y²] dx.

1. Basically you square the function if you're finding the volume in this case - it is y =x²+2. .:. y² = x^4 + 4x²+4

V = pi int [x^4 + 4x²+4] 2->0

V = pi [ x^5/5 + 4x^3/3 + 4x] 2->0

V = pi [ ( 32/5 + 32/3 + 8 ) - 0+0+0 ]

Therefore, the volume is pi [ 367/15 ].

2. Same rules apply. Make y the subject of x+3y-1 = 0, square the y function, integrate that and place the boundaries in your integration.

 

[Avenger6]

Ok thanks for the help. I am still a little confused with the second questions however. When I make y the subject and make it y² I get y²=(x-1/-3)², is this correct? Would I then intergrate to get (x-1/3)³ over 3???

 

[tommykins]

x+3y-1=0 from x = 0 > 8

y = (1-x)/3
y² = [ (1-x)/3 ]²
= [ (1-2x+x²/9 ]

V = pi int [ (1-2x+x²/9 ] 8->0
but you can take the /9 out to make it become
V = pi/9 int [ 1-2x+x² ] 8->0

V = pi/9 [ x - x² + x³/3 ]8->0
V = pi/9 [ (8- 64 + 512/3) - (0-0+0) ]

.:. V = 344pi/27 .

 

[Avenger6]

Thanks for all the help, made alot of sense .