How do you find the volume of the solid in the first octant that is bounded by the cylinder given by z = 16x^2 and the plane y = 5?

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Finding the Volume of the Solid

To determine the volume of the solid in the first octant bounded by the cylinder defined by z = 16x^2 and the plane y = 5, we first need to visualize the region we are working with. The solid is present in the first octant, where x, y, and z are all greater than or equal to zero.

1. Understand the Boundaries

The cylinder z = 16x^2 opens upwards and is symmetrical around the z-axis. The height of the solid at any slice parallel to the xy-plane is given by the surface of the cylinder. The plane y = 5 defines an upper boundary for our solid.

2. Set Up the Integral

We can express the volume V of the solid using a double integral over the xy-plane:

V = ∫∫_D z \, dA

where D is the region in the xy-plane defined by the limits of x and y.

3. Determine the Limits of Integration

In the first octant, we will have:

  • 0 <= x <= sqrt(5) (since y max is 5: y = 16x^2 gives max when z equals 0 at maximum area of y = 5)
  • 0 <= y <= 5
  • 0 <= z <= 16x^2

4. Volume Expression

The volume integral can be set up as follows:

V = ∫_0^5 ∫_0^{sqrt(y/16)} (16x^2) \, dx \, dy

5. Calculate the Inner Integral

Now we find the inner integral with respect to x:

∫_0^{sqrt(y/16)} (16x^2) \, dx = [16 * (1/3)x^3] | _0^{sqrt(y/16)} = (16/3)(y/16)^{3/2} = (y^{3/2}) / (12)

6. Calculate the Outer Integral

Next, substituting this result into the outer integral gives us:

V = ∫_0^5 (y^{3/2}) / 12 \, dy

Calculate:

V = (1/12) * [ (2/5)*(y^{5/2})] | _0^5 = (1/12)(2/5)(5^{5/2}) = (1/12)(2/5)(25 * sqrt(5)) = (5 * sqrt(5))/12

7. Final Volume Result

The volume of the solid bounded by the cylinder and the plane in the first octant is thus:

V = (5 * sqrt(5))/12

This provides a complete solution for finding the volume of the solid in the specified region.

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