Volume by Integration

  • #1
Find the volume of the solid obtained by rotating the region bounded by the curves:

x = 2 sqrt(y), x = 0, y = 9; about the y-axis

I have this graphed and everything so I'm not sure there's a need to worry about that.

Setup (integrating with respect to y):
Outer radius = x = 2 sqrt(y). Inner radius = x = 0. Bounded by y = 9 and 0.

9
∫π[2 sqrt(y)]^2 dy
0

Which is equivalent to:
_______9
(2πy^2)| = 36π
_______0

However, my book says the answer is 162π. What am I doing wrong?
 

Answers and Replies

  • #2
STEMucator
Homework Helper
2,075
140
Ahhh now you might've missed this :

x = 2[itex]\sqrt{y}[/itex]

So your volume integral becomes :

[itex]\pi \int_{0}^{9} 4y \space dy[/itex]

Since [itex]f^2(y) = 4y[/itex]
 
Last edited:
  • #3
Ahhh now you might've missed this :

x = 2[itex]\sqrt{y}[/itex]

So your volume integral becomes :

[itex]\int_{0}^{9} 4y \space dy[/itex]

Since [itex]f^2(y) = 4y[/itex]
That doesn't change what I get at the end. I'll still factor out the 4 and the π and what's left in the integral will still come to be (y^2) / 2. Multiply that by what's outside the integral and I still have 2πy^2.
 
  • #4
STEMucator
Homework Helper
2,075
140
That doesn't change what I get at the end. I'll still factor out the 4 and the π and what's left in the integral will still come to be (y^2) / 2. Multiply that by what's outside the integral and I still have 2πy^2.
Oh whoops, in my last post i meant your integral is really :

[itex]π \int_{0}^{9} 4y \space dy[/itex]

So, integrating it you get :

[itex]π \int_{0}^{9} 4y \space dy = 4π \int_{0}^{9} y \space dy = 2π[y^2]_{0}^{9} \space dy[/itex]

Do you see it now?
 
  • #5
Aah. I know what I did. I just thought that came out to 36π for some stupid reason. Alright thanks. Sorry about that. I'm a bit of a dope : D
 

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