Question

Find the volume of the region between the planes x + y + z = 4 and 4x + 4y + z =16 in the first octant.

Answer 1

You have \(x + y + z = 4\) and \(4x + 4y + z =16\)

ie, in 1st octant: \(x,y > 0\)

So \(z_1 = 4 - x - y\) and \(z_2 =16 - 4 x - 4 y = 4 (4 - x - y)\)

Let \(\tilde z = z_2 - z_1 =3(4 - x - y)\)

Volume = \(\int_A \tilde z (x,y) ~ dA\)

\(= 3 \int_{y = 0}^{4} \ \int_{x = 0}^{4 - y} ~ 4 - x - y ~ dx ~ dy\)

\(= 3 \int_{x = 0}^{4} \ \int_{y = 0}^{4 - x} ~ 4 - x - y ~ dy ~ dx \)

\(= 32\)

Answer 2

thank you so much c: I think I had the limits of integration set up wrong or something