Volume of Area's over Axis [Calculus Help Needed]

[MrCodeDude]

I DO NOT HAVE ANSWERS, WHICH IS WHY I NEED YOUR GUY'S HELP!

Find the volume rotated over the plane region bounded by the equations:

y = root [x], y=2, x=0

a) the x-axis
b) the line y = 2
c) the y-axis
d) the line x=-1

PART A
pi * integral from 0 to 4 [sqrt[x]^2]
= 8pi

PART B
pi * integral from 0 to 4 [(2 - sqrt[x])^2]
= 8pi / 3

PART C
pi * integral from 0 to 2 [(4-y^2)^2]
= 64pi / 15 <-- I think this is wrong

PART D
I need help on this one

 

[opticalmace]

i'm glad i'm not taking calculus anymore


--yes, i'm that helpful

 

[MrCodeDude]

 

[MrCodeDude]

BUMP

 

[MrCodeDude]

BUMP

 

[esun]

Just use the formula (for part D) V = integral of pi*R^2*dy evaluated from 0 to 2.

You just need to figure out what R is in this case. Look at a graph of the function (it's the same as rotating y = x^2 about y = -1). Since you have a dy anyway, you will need the function in terms of y, so you get x = y^2.

So y^2 will give you the distance from x = 0 to the function. Since you are rotating about x = -1, add one to this to find R. Therefore, R = y^2 + 1

So, the solution should be the integral of pi * (y^2 + 1)^2 * dy evaluated from 0 to 2. My TI-89 says that is 206*pi/15.

EDIT: Part C should be the same integral but R = y^2 only (I don't see where you got the 4 - y^2 from).