Finding Volume of Solid Bounded by Paraboloid and Planes

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SUMMARY

The volume of the solid enclosed by the paraboloid defined by the equation z = x^2 + 3y^2 and the planes x = 0, y = x, y = 1, and z = 0 can be calculated using double integration. The integration bounds for y are from 0 to 1, and for x from 0 to y. The correct setup for the double integral is ∫∫ (x^2 + 3y^2) dy dx, which simplifies to the evaluation of the integral to find the volume. A sketch of the solid and the region in the x-y plane is essential for determining the correct limits of integration.

PREREQUISITES
  • Understanding of double integrals in calculus
  • Familiarity with the concept of volume under surfaces
  • Knowledge of paraboloids and their equations
  • Ability to sketch regions in the x-y plane
NEXT STEPS
  • Learn about setting up double integrals for volume calculations
  • Study the properties and applications of paraboloids in three-dimensional geometry
  • Explore software tools for visualizing three-dimensional solids and their projections
  • Review examples of volume calculations involving multiple planes and surfaces
USEFUL FOR

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

Krayfish
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Homework Statement


Find the volume of the solid enclosed by the paraboloid z=x^2 + 3y^2 and the planes x=0, y=x, y=1, z=0

Homework Equations


I'm not really sure what's getting me about this, but I'm not really sure how to proceed after finding the x, y, and z intercepts...

The Attempt at a Solution


x intercept: 0
y intercept: 0
z intercept: 0

Would I just take bounds of y to be 1 and x and the bounds of x to be 0 and 1?
∫∫x^2 + 3y^2 dydx → The dy result would be yx^2 + y^3 → (x^3+x^3 -(x^2+1)) → 2X^3 - x^2 - 1 and then continue to integrate... ∫2x^3-x^2-1 dx → ((x^4)/2 -(x^2)/3 -x) → 1/2 - 1/3 - 1 = -5/6?

If that correct or am I missing something important?
 
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Krayfish said:

Homework Statement


Find the volume of the solid enclosed by the paraboloid z=x^2 + 3y^2 and the planes x=0, y=x, y=1, z=0

Homework Equations


I'm not really sure what's getting me about this, but I'm not really sure how to proceed after finding the x, y, and z intercepts...

The Attempt at a Solution


x intercept: 0
y intercept: 0
z intercept: 0

Would I just take bounds of y to be 1 and x and the bounds of x to be 0 and 1?
∫∫x^2 + 3y^2 dydx → The dy result would be yx^2 + y^3 → (x^3+x^3 -(x^2+1)) → 2X^3 - x^2 - 1 and then continue to integrate... ∫2x^3-x^2-1 dx → ((x^4)/2 -(x^2)/3 -x) → 1/2 - 1/3 - 1 = -5/6?

If that correct or am I missing something important?
Have you drawn a sketch of the solid whose volume you're supposed to find? Also, you should draw a sketch of the region in the x-y plane that forms the base of this solid. That would help you figure out what the integration limits need to be.
 
Mark44 said:
Have you drawn a sketch of the solid whose volume you're supposed to find? Also, you should draw a sketch of the region in the x-y plane that forms the base of this solid. That would help you figure out what the integration limits need to be.
I struggled to sketch it myself so I ended up using a software and plotted the parabaloid with the two y= planes. I'm not really sure where to go from there, all the examples we've done thus far have had at least 1 non zero intercept.
 

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