Is My Solution for the Triple Integral Correct?

In summary, the triple integral evaluation for \int_{0}^{1} \int_{x^2}^{1} \int_{0}^{3y} ({y+2x^2z})dz dy dx, after integrating and plugging in the limits, simplifies to \frac{32}{21}. This was verified by multiple methods and took approximately 3 minutes to solve.
  • #1
mathzeroh
98
0
Triple Integral Evaluation (quick and easy)

Homework Statement


[itex]
\int_{0}^{1} \int_{x^2}^{1} \int_{0}^{3y} ({y+2x^2z})dz dy dx
[/itex]


Homework Equations



None.

The Attempt at a Solution



Here is what I got at the end (the LaTeX takes too long to code in here, plus its not showing up):

59/36 because after integrating the whole thing, and then putting in the very last limits (0 and 1), all of the x's go away leaving just the coefficients which I worked out to get 59/36...can someone please verify this for me? I've checked it twice and got the same solution..it takes about 3 minutes to do if you're a genious (unlike me, so I'm trying to appeal to the math geniuses)..

thanks!
 
Last edited:
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  • #2
I get (actually Maple) [itex] \frac{32}{21} [/itex]

∫{0}^{3y}(y+2x²z)dz= 9x²y²+3y²

∫_{x²}^{1}( 9x²y²+3y²)dy= -3x^{8}-x^{6}+3x²+1



∫_{0}^{1}(-3x^{8}-x^{6}+3x²+1) dx= ((32)/(21))
 
  • #3
did it by hand and got 32/21
 
  • #4
I did it by hand, too. :-p I had to type everything and then use the "ctrl c-ctrl v" keys. It's not as easy as it seems...:biggrin:

Daniel.
 
  • #5
[tex]\int_0^1\int_{x^2}^1\int_0^3y (y+ 2x^2z)dzdydx[/tex]
[tex]\int_0^1\int_{x^2}^1\left[yz+ x^2z^2\right]_0^{3y}dydx[/tex]
[tex]3\int_0^1\int_{x^2}^1(1+ x^2)y^2 dy dx[/tex]
[tex]\int_0^1\left[(1+3x^2)y^3\right]_{x^2}^1 dx[/tex]
[tex]\int_0^1(1+3x^2)(1-x^6)dx[/tex]
[tex]\int_0^1(-3x^8- x^6+ 3x^2+ 1)dx[/tex]
[tex]\left[-\frac{1}{3}x^9-\frac{1}{7}x^7+ x^3+ x\right]_0^1[/tex]
[tex]-\frac{1}{3}-\frac{1}{7}+ 1+ 1[/tex]
[tex]-\frac{7}{21}-\frac{3}{21}+\frac{21}{21}+\frac{21}{21}[/tex]
[tex]\frac{32}{21}[/tex]
 

Related to Is My Solution for the Triple Integral Correct?

What is a triple integral?

A triple integral is a mathematical tool used to calculate the volume of a three-dimensional region. It involves integrating a function over a three-dimensional space, with each integral corresponding to one of the three dimensions.

When is a triple integral used?

A triple integral is commonly used in physics and engineering to calculate physical quantities such as mass, center of mass, and moment of inertia. It is also used in multivariable calculus to solve problems involving three-dimensional shapes and volumes.

How is a triple integral evaluated?

A triple integral is evaluated by breaking down the three-dimensional region into smaller, simpler shapes, such as cubes or cylinders. These shapes are then integrated using standard techniques, such as the shell method or the disk method. The individual integrals are then combined to calculate the overall triple integral.

What are the limits of integration in a triple integral?

The limits of integration in a triple integral correspond to the boundaries of the three-dimensional region being integrated over. These boundaries can be defined by equations or inequalities in terms of the three variables, and they determine the range of values that each variable can take within the integral.

What are some practical applications of triple integrals?

Triple integrals have many practical applications in various fields, such as physics, engineering, and economics. They can be used to calculate the volume of three-dimensional objects, the flow of fluids, the distribution of electric charge, and the probability of events in a three-dimensional space, among others.

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