Finding Volume by use of Triple Integrals

In summary, the problem asks to find the volume enclosed by two elliptic paraboloids, one flipped on the xz-plane and moved up 8 units. The domain of this volume is given by the equation 4=x^{2}+z^{2}. The limits of integration are found to be x^{2}+z^{2}\leqy\leq8-x^{2}-z^{2}, -\sqrt{y-z^{2}}\leqx\leq\sqrt{y-z^{2}}, and -2\leqz\leq2. The integrand for the volume is 1. Polar coordinates can be used after the dy integral is done.
  • #1
maiad
102
0

Homework Statement


Find the Volume of the solid eclose by y=x[itex]^{2}[/itex]+z[itex]^{2}[/itex] and y=8-x[itex]^{2}[/itex]-z[itex]^{2}[/itex]


The Attempt at a Solution



Well know they're both elliptic paraboloids except one is flipped on the xz-plane and moved up 8 units. Knowing this, i equated the two equations and got 4=x[itex]^{x}[/itex]+z[itex]^{2}[/itex] which is the Domain of this volume.

I found the limits of the intergration... x[itex]^{2}[/itex]+z[itex]^{2}[/itex][itex]\leq[/itex]y[itex]\leq[/itex]8-x[itex]^{2}[/itex]-z[itex]^{2}[/itex]...-[itex]\sqrt{y-z^{2}}[/itex][itex]\leq[/itex]x[itex]\leq[/itex][itex]\sqrt{y-z^{2}}[/itex]...-2[itex]\leq[/itex]z[itex]\leq[/itex]2

But my problem now is what am i actually integrating. would it be x[itex]^{2}[/itex]+z[itex]^{2}[/itex]?
 
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  • #2
maiad said:

Homework Statement


Find the Volume of the solid eclose by y=x[itex]^{2}[/itex]+z[itex]^{2}[/itex] and y=8-x[itex]^{2}[/itex]-z[itex]^{2}[/itex]


The Attempt at a Solution



Well know they're both elliptic paraboloids except one is flipped on the xz-plane and moved up 8 units. Knowing this, i equated the two equations and got 4=x[itex]^{x}[/itex]+z[itex]^{2}[/itex] which is the Domain of this volume.

I found the limits of the intergration... x[itex]^{2}[/itex]+z[itex]^{2}[/itex][itex]\leq[/itex]y[itex]\leq[/itex]8-x[itex]^{2}[/itex]-z[itex]^{2}[/itex]...-[itex]\sqrt{y-z^{2}}[/itex][itex]\leq[/itex]x[itex]\leq[/itex][itex]\sqrt{y-z^{2}}[/itex]...-2[itex]\leq[/itex]z[itex]\leq[/itex]2

But my problem now is what am i actually integrating. would it be x[itex]^{2}[/itex]+z[itex]^{2}[/itex]?

No. The integrand would be ##1##.$$
Vol = \iiint_R 1\, dV$$And I would suggest you using polar coordinates in the ##xz## plane after you do the ##dy## integral.
 

Related to Finding Volume by use of Triple Integrals

What is the concept of triple integrals?

Triple integrals are a mathematical concept used to find the volume of a three-dimensional object or region. They involve calculating the integral of a function over a three-dimensional space.

How do you set up a triple integral?

To set up a triple integral, you must first determine the boundaries for each variable (x, y, z) in the given three-dimensional space. These boundaries can be found by visualizing the object or by using equations that define the surface of the object. Once the boundaries are determined, the integral is written as a nested integral with the innermost integral representing the first variable, followed by the second variable, and finally the third variable.

What are the applications of triple integrals?

Triple integrals have various applications in fields such as physics, engineering, and economics. They are used to calculate the volume of three-dimensional objects, mass and density of objects, and to find the center of mass of a three-dimensional object. They are also used in solving optimization problems and in calculating the probability of events in three-dimensional space.

What are the limitations of using triple integrals to find volume?

While triple integrals are useful in finding the volume of complex three-dimensional objects, they have limitations in certain cases. They can only be used for finding the volume of objects with smooth surfaces and for objects that can be described by equations. Additionally, the process of setting up and solving triple integrals can be time-consuming and complex.

How do triple integrals relate to double integrals?

Triple integrals are an extension of double integrals, which are used to find the area of two-dimensional objects. While double integrals integrate over a two-dimensional region, triple integrals integrate over a three-dimensional region. They both use similar concepts and techniques, such as determining boundaries and using different coordinate systems, but triple integrals involve an additional variable and more complex calculations.

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