Triple Integral question

  • Thread starter Redoctober
  • Start date
  • #1
48
1
I need some help in this

The question states find the volume for the region bounded below by z = 3 - 2y and above by z = x^2 + y^2 using cylindrical coordinates .

Now i tried to do it but i got stuck with the part where i find the radius . i found
r^2 = 3-2r*sin(zeta) :( its hard to factorize and use this for the boundary

i already know that z will range from r^2 to 3-2r*sin(zeta) and zeta range from zero to 2 pie

so how can this question be solved ??? :)

btw the correct answer is 8pie
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
961
First, you have "below" and "above" reversed. The plane z= 3- 2y is above the paraboloid [itex]z= x^2+ y^2[/itex] in the bounded region.

They intersect where [itex]z= x^2+ y^2= 3- 3y[/itex] which is equivalent to [itex]x^2+ y^2+ 3y= 3[/itex]. [itex]x^2+ y^2+ 3y+ 9/4= x^2+ (y+3/2)^2= 3+ 9/4= 21/4[/itex]. That is a circle with center at (0, -3/2) which is why ordinary polar coordinates do not give a simple equation. Either
1: integrate with x from [itex]-\sqrt{21}/2[/itex] to [itex]\sqrt{21}/2[/itex] and, for each x, y from [itex]-3/2- \sqrt{21/4- x^2}[/itex] to [itex]-3/2+ \sqrt{21/4- x^2}[/itex].

2: Let [itex]x= rcos(\theta)[/itex] and [itex]y= -3/2+ rsin(\theta)[/itex], shifting the origin to (0, -3/2).
 
  • #3
48
1
Oh i c :D !!! Thanks alot . I wasn't aware of shifting the origin :) !!

Thanks :D !
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
961
For desert?
 

Related Threads on Triple Integral question

  • Last Post
Replies
7
Views
4K
Replies
3
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
5
Views
7K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
2K
Top