Double integral volume problem

Click For Summary
SUMMARY

The discussion centers on calculating the volume of the solid beneath the plane defined by the equation z = 4x and above the circular region described by x² + y² = 16 in the xy-plane. Participants clarified that part of the plane z = 4x does indeed lie above the xy-plane, indicating that a solid exists between the two surfaces. The correct approach involves using double integrals and polar coordinates to evaluate the volume, rather than concluding that the volume is zero.

PREREQUISITES
  • Understanding of double integrals in calculus
  • Familiarity with polar coordinates
  • Knowledge of volume calculation beneath surfaces
  • Ability to sketch and interpret 3D graphs
NEXT STEPS
  • Study the application of double integrals in volume calculations
  • Learn how to convert Cartesian coordinates to polar coordinates
  • Explore the use of triple integrals for volume under surfaces
  • Practice sketching 3D surfaces and their intersections
USEFUL FOR

Students studying calculus, particularly those focusing on multivariable calculus and volume calculations, as well as educators looking for examples of integrating over circular regions.

nate9519
Messages
47
Reaction score
0

Homework Statement


find the volume of the solid below the plane z = 4x and above the circle x^2 + y^2 = 16 in the xy plane

Homework Equations

The Attempt at a Solution


This totally confused me. I didn't think the plane z = 4x sat above the xy plane. If that is true then there would be no solid between the two graphs. I ended up putting zero for the answer. was I right or wrong
 
Physics news on Phys.org
nate9519 said:

Homework Statement


find the volume of the solid below the plane z = 4x and above the circle x^2 + y^2 = 16 in the xy plane

Homework Equations

The Attempt at a Solution


This totally confused me. I didn't think the plane z = 4x sat above the xy plane.
Part of the z = 4x plane lies above the xy plane. Did you draw a sketch of the plane and the circle?
nate9519 said:
If that is true then there would be no solid between the two graphs. I ended up putting zero for the answer. was I right or wrong
 
$$\iint_R z \space dA = \iiint_V \space dV$$

$$\iiint_V \space dV = \iint_R \int_0^z \space dzdA$$

;)

Polar co-ordinates.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
2K
  • · Replies 27 ·
Replies
27
Views
3K