Triple Integral, already solved, need checked

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SUMMARY

The volume of the region S in the first octant under the plane defined by the equation 3x + 2y + z = 4 is calculated using a triple integral. The correct setup for the integral is \int^{\frac{4}{3}}_{0}\int^{2-\frac{3}{2}x}_{0}\int^{4-3x-2y}_{0}dzdydx, which evaluates to \frac{448}{27}. A key point of discussion was the correction of the y-bound in the integral setup, which was initially misrepresented.

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  • Understanding of triple integrals in calculus
  • Familiarity with the concept of regions in three-dimensional space
  • Knowledge of the first octant and its boundaries
  • Ability to manipulate and evaluate integrals
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Homework Statement


Let S be the region in the first octant under the plane 3x + 2y +z = 4. Find the volume of S.


Homework Equations


idk?

The Attempt at a Solution



\int^{\frac{4}{3}}_{0}\int^{\frac{3}{2}x + 2}_{0}\int^{-3x - 2y + 4}_{0}dzdydx

= \int^{\frac{4}{3}}_{0}\int^{\frac{3}{2}x + 2}_{0}(-3x - 2y + 4)dydx

= \int^{\frac{4}{3}}_{0}(-\frac{15}{2}x^{2} - 6x + 4)dx

= 448/27
 
Last edited:
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Sorry, took a little while to figure out the symbols and dealies, pretty sure it's good to go right now.
 
Mumble, mumble..gotta check the bounds:

The upper line in the z=0 plane obeys the equation 3x+2y=4

Thus,
0\leq{x}\leq\frac{4}{3}, 0\leq{y}\leq{2}-\frac{3}{2}x
Thus, you've got a sign wrong in the y-bound, according to my view.
 
arildno said:
Mumble, mumble..gotta check the bounds:

The upper line in the z=0 plane obeys the equation 3x+2y=4

Thus,
0\leq{x}\leq\frac{4}{3}, 0\leq{y}\leq{2}-\frac{3}{2}x
Thus, you've got a sign wrong in the y-bound, according to my view.

You're so right, thankyou!
 

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