Volume using intergals

  • Thread starter flash9286
  • Start date
  • #1
9
0

Main Question or Discussion Point

If the region bounded by y=9-x^2,y=0,x=0. Is rotated about the x axis the what would the volume be. I'm got 407. For the same region rotate about the y=9, and i got 153. Can someone check these answers i think they are wrong.
 

Answers and Replies

  • #2
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
131
Set up the equations you used! :smile:
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,805
932
You know, I could have sworn that when you rotate something a "[itex]\pi[/itex]" would show up!

How about showing us exactly what you integrated and how you got those answers?
 
  • #4
9
0
The first integral would be, I think: b=3,a=0, on the integral pi*(9-x^2)^2


The second one, which I'm not sure about, would be same b and a, on the integral pi*(9-(9-x^2))^2
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,805
932
The first is set up correctly but that surely is not "407"!

When you rotate around the y-axis, your "disks" will be moving up the y axis. The radius of each disk will be x as a function of y and their "thickness" is dy.
 
  • #6
9
0
Thats what i get when I put it into my ti-83. I'm pretty sure that the first one is right. But I'm having a hard time imaging the shape you get when you rotate the area around y=9.
 
  • #7
9
0
[tex] \int_{0}^{3} pi*9^2-(pi*(9-9+x^2)^ dx [/tex] is what the second integral maybe?
 
  • #8
9
0
Never mind guys, I just checked my work on maple and the answers I had first were right, I should stop second guessing my self :),
 

Related Threads on Volume using intergals

Replies
6
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
12
Views
3K
  • Last Post
Replies
7
Views
5K
  • Last Post
Replies
1
Views
965
  • Last Post
Replies
5
Views
2K
Top