find the volume bounded by the parabolic cylinder z=4-x^2 and the planes x=0, y=0, y=6 and z=0
I just want some help figuring out the limits of this question :
So for z, we have to integrate from z=0 to z=4-x^2 (am i right?)
W.r.t. y, we should integrate from y=0 y=6 (am i right?)
w.r.t. x , can you please explain?
Thanks!
The volume we want is in the 1st octant, since all the coordinate planes form part of the boundary. The cylinder is sliced off at y=6
So the volume is just 6 times the area under the curve z = 4-x^2
v = 6∫[0,2] 4-x^2 dx
If you insist on doing a triple integral, then you are correct as far as you go. The last step is just to integrate x from 0 to 2, since we want z=0 on the parabola.
You are correct about the limits for z and y. The integration limits for x depend on the shape of the region bounded by the parabolic cylinder and the given planes.
To determine the limits for x, we need to examine the intersection points of the parabolic cylinder and the planes x=0 and x=6.
At x=0, the equation of the parabolic cylinder becomes z=4-0^2=4. Therefore, at x=0, the parabolic cylinder intersects the plane z=0 (the xy-plane) at z=0 and at z=4.
At x=6, the equation of the parabolic cylinder becomes z=4-6^2=-32. Therefore, at x=6, the parabolic cylinder does not intersect the plane z=0 (the xy-plane).
So, the appropriate limits for x will be from x=0 to x=a, where 'a' is the x-coordinate of the intersection point between the parabolic cylinder and the plane z=0.
To find 'a', we set z=0 in the equation of the parabolic cylinder: 0=4-x^2. Solving for x, we get x=±2.
However, we need to consider only the positive value of x since the negative value corresponds to the other side of the parabolic cylinder.
Therefore, the limits for x will be from x=0 to x=2.
In summary, to find the volume bounded by the parabolic cylinder z=4-x^2 and the planes x=0, y=0, y=6, and z=0, you would integrate with respect to x from x=0 to x=2, with respect to y from y=0 to y=6, and with respect to z from z=0 to z=4-x^2.
I hope this clarifies the limits of integration for you!
Yes, you are correct about the limits of integration for z and y. Now let's determine the limits of integration for x.
To find the limits of integration for x, we need to consider the intersection points of the parabolic cylinder z = 4 - x^2 with the planes x = 0 and y = 6.
First, let's consider the plane x = 0. When x = 0, the equation of the parabolic cylinder becomes z = 4 - 0^2 = 4. So, the intersection point with the plane x = 0 is (0, 0, 4).
Now, let's consider the plane y = 6. In this case, we need to find the intersection points between the parabolic cylinder and this plane. By substituting y = 6 into the equation of the parabolic cylinder, we get:
z = 4 - x^2.
Now, let's set z to 0 and solve for x:
0 = 4 - x^2.
Rearranging this equation, we have:
x^2 = 4.
Taking the square root on both sides, we get:
x = ±2.
So, the intersection points with the plane y = 6 are (-2, 6, 0) and (2, 6, 0).
Therefore, the limits of integration for x are from -2 to 2, because the parabolic cylinder intersects the plane y = 6 at x = -2 and x = 2.
To summarize:
x limits: -2 to 2
y limits: 0 to 6
z limits: 0 to 4 - x^2
Now you can set up the triple integral to find the volume bounded by the parabolic cylinder and the given planes using these limits of integration.