set up all six possible triple integrals of the tetrahedron enclosed by the coordinate planes and the plane x + 2y + 3z = 30

Question

tkhunny · Answer

Divide by 30, creating the "Intercept Form", and it should be relatively easy.

$\dfrac{x}{30} + \dfrac{y}{15} + \dfrac{z}{10} = 1$

anonymous · Answer

im trying to post a picture of my paper but apparently its not working

anonymous · Answer

$$\int\limits_{0}^{10} \int\limits_{0}^{3z + 15}  \int\limits_{0}^{2y -30}   dx dy dz$$

anonymous · Answer

$$\int\limits_{0}^{15} \int\limits_{0}^{(2/3)y + 10} \int\limits_{0}^{(z-10)/3} dx dz dy$$

anonymous · Answer

check the first two please!

anonymous · Answer

$$\int\limits_{0}^{30} \int\limits_{0}^{1/2(x) +15} \int\limits_{0}^{2/3 (y) + 10}  dz dy dx$$

tkhunny · Answer

No good.  The most interior limits must have two variables,

x:[0,30 - 2y - 3z]
y: [0,15 - (3/2)z]
z: [0,10]

x:[0,30 - 2y - 3z]
z: [0,10 - (2/3)z]
y: [0,15]

etc.