Question

Set up the triple Integral SI V over the tetrahedron formed by the coordinate planes and the plane X + y + 24 using six different orders of Integration de dy dx dx dy dy do dy dx de ar ddy dr dy (b) Find the volume of the tetrahedron.

Answer 1

Explanation:

For Part A:

Refer to the photo for an explanation on how to graph z=x+y+24 and how to set up the first integral. I will not explain how to do the other five but I have placed them beneath....

`\int\limits^(24)_0 {} \,\int\limits^(z-24)_0 {} \,\int\limits^(z-y-24)_0 {} \, 1dxdydz \\\\`

Where limits are,
`0\leq x\leq z-y-24\\0\leq y\leq z-24\\0\leq z\leq 24`

`\int\limits^(-24)_0 {} \,\int\limits^(y+24)_0 {} \,\int\limits^(z-y-24)_0 {} \, 1dxdzdy\\\\`

Where limits are,
`0\leq x\leq z-y-24\\0\leq z\leq y+24\\0\leq y\leq -24`

`\int\limits^(24)_0 {} \,\int\limits^(z-24)_0 {} \,\int\limits^(z-x-24)_0 {} \, 1dydxdz\\\\`

Where limits are,
`0\leq y\leq z-x-24\\0\leq x\leq z-24\\0\leq z\leq 24`

`\int\limits^(-24)_0 {} \,\int\limits^(x+24)_0 {} \,\int\limits^(z-x-24)_0 {} \, 1dydzdx\\\\`

Where limits are,
`0\leq y\leq z-x-24\\0\leq z\leq x+24\\0\leq x\leq -24`

`\int\limits^(-24)_0 {} \,\int\limits^(-y-24)_0 {} \,\int\limits^(x+y+24)_0 {} \, 1dzdxdy\\\\`

Where limits are,
`0\leq z\leq x+y+24\\0\leq x\leq -y-24\\0\leq y\leq -24`

`\int\limits^(-24)_0 {} \,\int\limits^(-x-24)_0 {} \,\int\limits^(x+y+24)_0 {} \, 1dzdydx`

Where Limits are,
`0\leq z\leq x+y+24\\0\leq y\leq -x-24\\0\leq x\leq -24`

For Part B:

All the above limits equal the same volume. So I will just solve for one of the limits.

Evaluate,
`\int\limits^(24)_0 {} \,\int\limits^(z-24)_0 {} \,\int\limits^(z-y-24)_0 {} \, 1dxdydz \\\\`

`\int\limits^(24)_0 {} \,\int\limits^(z-24)_0 {} \, (x]^(z-y-24) _0){} dydz \\\\`

`\int\limits^(24)_0 {} \,\int\limits^(z-24)_0 {} \, [(z-y-24)-0]{} dydz \\\\`

`\int\limits^(24)_0 {} \,\int\limits^(z-24)_0 {} \,z-y-24{} dydz \\\\`

`\int\limits^(24)_0 {} \,[0-(y^(2) )/(2) -0]{} dz \\\\`

`\int\limits^(24)_0 {} \,(y^(2) )/(2) ]^(z-24) _(0) {} dz \\\\`

`\int\limits^(24)_0 {} \,[(((z-24)^(2))/(2))-0] {} dz \\\\`

`\int\limits^(24)_0 {} \,[(z^2)/(2)-288] {} dz \\\\`

`[(z^3)/(6)-288z]^(24) _(0) {} \\\\`

`[((24^3)/(6)-288(24))-0] {}`

`(4608-6912)=-2304\\`

Answer: -2304