Finding Volume of Solid Bounded by Paraboloid and Planes

In summary, the task is to find the volume of a solid enclosed by a paraboloid and three planes. The discussion includes the attempt at a solution which involves finding the x, y, and z intercepts and integrating using the integration limits of 0 and 1. However, the person also mentions struggling with sketching the solid and determining the integration limits.
  • #1
Krayfish
4
0

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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  • #2
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.
 
  • #3
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.
 

What is double integration?

Double integration is a mathematical process used to calculate the volume under a two-dimensional curve. It involves integrating the function twice with respect to two different variables.

What are the types of double integration?

The two types of double integration are iterated integration and polar integration. In iterated integration, the limits of integration are given for the two variables separately. In polar integration, the limits are given in terms of polar coordinates.

What is the purpose of bounds in double integration?

Bounds in double integration specify the range over which the integration is performed. They help define the area or volume that is being calculated and ensure that the integration is accurate.

How is double integration used in physics?

In physics, double integration is used to calculate the center of mass, moments of inertia, and work done by a force. It is also used in electromagnetism to calculate electric and magnetic fields.

What are some applications of double integration?

Some common applications of double integration include calculating the area under a curve, finding the volume of a solid, and determining the average value of a function over a given area. It is also used in fields such as engineering, economics, and statistics.

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