3. if ( i ) is the integral: ( iiint_{t} z d v ) where ( t ) is the tetrahedron consisting of all of the…

3. if ( i ) is the integral: ( iiint_{t} z d v ) where ( t ) is the tetrahedron consisting of all of the points in the first octant such that ( x + 3 y + 5 z leq 15 ), then express ( i ) in all of the 6 possible orders of integration. you do not need to compute this integral!

3. if ( i ) is the integral: ( iiint_{t} z d v ) where ( t ) is the tetrahedron consisting of all of the points in the first octant such that ( x + 3 y + 5 z leq 15 ), then express ( i ) in all of the 6 possible orders of integration. you do not need to compute this integral!

Answer

Explanation:

Step1: Determine the limits for (z)

From (x + 3y+5z\leq15), we can express (z) as (z\leq\frac{15 - x - 3y}{5}). Since we are in the first - octant ((x\geq0,y\geq0,z\geq0)), the lower limit for (z) is (z = 0) and the upper limit is (z=\frac{15 - x - 3y}{5}).

Step2: Determine the limits for (y)

Set (z = 0) in (x + 3y+5z\leq15), we get (3y\leq15 - x) or (y\leq\frac{15 - x}{3}). The lower limit for (y) is (y = 0) and the upper limit is (y=\frac{15 - x}{3}) (with (x\geq0)).

Step3: Determine the limits for (x)

Set (y = z=0) in (x + 3y+5z\leq15), we get (x\leq15). The lower limit for (x) is (x = 0) and the upper limit is (x = 15).

So one order of integration is (\int_{x = 0}^{15}\int_{y = 0}^{\frac{15 - x}{3}}\int_{z = 0}^{\frac{15 - x - 3y}{5}}z\ dV)

Another order:

Step1: Determine the limits for (y)

From (x + 3y+5z\leq15), we can express (y) as (y\leq\frac{15 - x - 5z}{3}). Lower limit (y = 0)

Step2: Determine the limits for (x)

Set (y = 0), then (x\leq15 - 5z). Lower limit (x = 0)

Step3: Determine the limits for (z)

Set (x=y = 0), then (z\leq3). Lower limit (z = 0)

So another order is (\int_{z = 0}^{3}\int_{x = 0}^{15 - 5z}\int_{y = 0}^{\frac{15 - x - 5z}{3}}z\ dV)

We can also change the order of (x) and (y) in the first - derived triple - integral.

If we first integrate with respect to (x):

Step1: Determine the limits for (x)

From (x+3y + 5z\leq15), we have (x\leq15 - 3y - 5z). Lower limit (x = 0)

Step2: Determine the limits for (y)

Set (x = 0), then (3y\leq15 - 5z) or (y\leq\frac{15 - 5z}{3}). Lower limit (y = 0)

Step3: Determine the limits for (z)

Set (x=y = 0), (z\leq3). Lower limit (z = 0)

So (\int_{z = 0}^{3}\int_{y = 0}^{\frac{15 - 5z}{3}}\int_{x = 0}^{15 - 3y - 5z}z\ dV)

Answer:

The six possible orders of integration are:

  1. (\int_{x = 0}^{15}\int_{y = 0}^{\frac{15 - x}{3}}\int_{z = 0}^{\frac{15 - x - 3y}{5}}z\ dV)
  2. (\int_{x = 0}^{15}\int_{z = 0}^{\frac{15 - x}{5}}\int_{y = 0}^{\frac{15 - x - 5z}{3}}z\ dV)
  3. (\int_{y = 0}^{5}\int_{x = 0}^{15 - 3y}\int_{z = 0}^{\frac{15 - x - 3y}{5}}z\ dV)
  4. (\int_{y = 0}^{5}\int_{z = 0}^{\frac{15 - 3y}{5}}\int_{x = 0}^{15 - 3y - 5z}z\ dV)
  5. (\int_{z = 0}^{3}\int_{x = 0}^{15 - 5z}\int_{y = 0}^{\frac{15 - x - 5z}{3}}z\ dV)
  6. (\int_{z = 0}^{3}\int_{y = 0}^{\frac{15 - 5z}{3}}\int_{x = 0}^{15 - 3y - 5z}z\ dV)