How do I find the volume using cylindrical coordinates for the given region?

In summary, the question is asking for the volume of a region bounded above by z = 3 - 2y and below by z = x^2 + y^2 using cylindrical coordinates. The solution involves finding the intersection of these two surfaces and using a shifted origin to simplify the integration process. The final answer is 8π.
  • #1
Redoctober
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
 
Physics news on Phys.org
  • #2
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
Oh i c :D ! Thanks a lot . I wasn't aware of shifting the origin :) !

Thanks :D !
 
  • #4
LOL 8pie
 
  • #5
For desert?
 

1. What is a triple integral?

A triple integral is an integral with three variables, used to calculate the volume of a three-dimensional region. It is represented by three nested integrals, with each variable representing a different dimension.

2. How do you set up a triple integral?

To set up a triple integral, you first need to identify the limits of integration for each variable, which are determined by the boundaries of the three-dimensional region. Then, you multiply the function being integrated by the differentials of each variable and integrate from the lower limit to the upper limit for each variable.

3. What is the purpose of a triple integral?

The purpose of a triple integral is to calculate the volume of a three-dimensional region. It is often used in physics and engineering to determine the mass, center of mass, and moment of inertia of an object.

4. What is the difference between a triple integral and a double integral?

A triple integral integrates a function over a three-dimensional region, while a double integral integrates a function over a two-dimensional region. The limits of integration and the number of variables also differ between the two.

5. What are some applications of triple integrals?

Triple integrals have various applications in physics, engineering, and mathematics. They are used to calculate the volume of solids, the mass and center of mass of objects, the moment of inertia of rigid bodies, and the probability of events in three-dimensional probability distributions.

Similar threads

Replies
5
Views
1K
Replies
4
Views
187
Replies
3
Views
530
Replies
4
Views
2K
  • Calculus
Replies
8
Views
2K
Replies
10
Views
2K
  • Calculus
Replies
5
Views
1K
Replies
3
Views
2K
Replies
1
Views
2K
  • Calculus
Replies
3
Views
885
Back
Top