Find the volume of the solid in the first octant bounded by the cylinder z = 16 – x² and the plane y = 5

To find the volume of the solid in the first octant bounded by the surface defined by the cylinder z = 16 – x² and the plane y = 5, we will set up a triple integral.

First, let’s identify the boundaries of the solid:

• The cylinder equation z = 16 – x² limits the height of the solid.
• The plane y = 5 restricts the extent in the y-direction.
• Since we are only considering the first octant, x, y, z ≥ 0.

Next, we need to find the limits for our integral. The variable x can range from 0 to the maximum value where the cylinder meets the plane:

• At y = 5, z can take the values from 0 to 16 – x².
• The maximum value for x happens when z = 0, so solving 16 – x² = 0 gives x = 4.

Therefore, the limits for x are [0, 4], for y are [0, 5], and for z are [0, 16 – x²].

The volume V can be calculated using the triple integral:

V = ∫[0 to 4] ∫[0 to 5] ∫[0 to 16 - x²] dz \, dy \, dx

Now, let’s compute the integral step by step:

V = ∫[0 to 4] ∫[0 to 5] (16 - x²) dy \, dx

Calculating the inner integral:

∫[0 to 5] (16 - x²) dy = (16 - x²) * y |[0 to 5] = 5(16 - x²)

Then, we have:

V = ∫[0 to 4] 5(16 - x²) dx

Now, compute this integral:

V = 5 ∫[0 to 4] (16 - x²) dx

This can be simplified further:

∫ (16 - x²) dx = 16x - (x³/3) |[0 to 4]

Evaluating from 0 to 4 gives us:

16(4) - (4³/3) = 64 - (64/3) = 64/3

Thus, substituting back:

V = 5 * (64/3) = 320/3

The volume of the solid is V = 320/3 cubic units.