Question

Find the surface area of the portion S of the cone z2=x2+y2, where z≥0, contained within the cylinder y2+z2≤36.

Answer 1

To find the surface area of the cone
`z^(2) =x^(2) +y^(2)` truncated by the cylinder
`y^(2)+z^(2)`≤36, we set the equations equal and find a circle of radius 6 in the yz-plane. Using double integrals in cylindrical coordinates with the surface element as dA, we calculate the surface area by integrating over the radius and angle.

Step-by-step explanation:

The student is tasked with finding the surface area of a specific portion of a cone defined by the equation
`z^(2)= x^(2) +y^(2)`, which is truncated by a cylinder with the constraint
`y^(2)+ z^(2)` ≤ 36. Since this is a cone with z ≥ 0, we are only considering the upper half of the cone, enclosed by the cylinder. To find the surface area of this truncated cone part S, we need to use double integrals in cylindrical coordinates.

Let's find the radius of the intersection circle of the cone and cylinder. Setting the equations equal to each other, substituting y from the cylinder's equation into the cone's equation, we have:

`(36 - z^(2)) +z^(2)` = 36

This gives us a circle of radius 6 in the yz-plane. Noting that the surface area element dA on a cone can be expressed in terms of the slant height, the double integral will take the form:

∫ ∫ sqrt(1 + r2) r dθ dr, where r is 0 to 6 and θ is 0 to 2π.

After evaluating this double integral, we would have the surface area of the portion S of the cone that falls within the cylinder.

Answer 2

The surface area S of the portion of the cone
`\( z^2 = x^2 + y^2 \)`, where
`\( z \geq 0 \)`, contained within the cylinder
`\( y^2 + z^2 \leq 36 \)` is
`\( 216\pi \)` square units.

Step-by-step explanation:

To find the surface area of the portion S of the cone
`\( z^2 = x^2 + y^2 \), where \( z \geq 0 \)`, contained within the cylinder
`\( y^2 + z^2 \leq 36 \)`, we'll first parameterize the surfaces of the cone and the cylinder.

Given the cone equation
`\( z^2 = x^2 + y^2 \)`, we can rewrite it in terms of cylindrical coordinates as z = r, where r represents the radial distance from the z-axis.

In cylindrical coordinates, the equation of the cylinder
`\( y^2 + z^2 = 36 \)`becomes
`\( r^2 = 36 \)`, or simply r = 6.

The surface area of the portion S of the cone can be obtained by integrating the surface area element over the region described by the cylindrical coordinates.

The surface area element dS for a surface given in cylindrical coordinates is
`\( dS = r \, dz \, dr \, d\theta \).`

We need to determine the limits of integration for r,
`\theta \)`, and z to cover the region of interest.

Limits of integration:

- r varies from 0 to 6 (since r = 6 for the cylinder).

-
`\( \theta \)` will vary from 0 to
`\( 2\pi \)` to cover the entire circle around the z-axis.

- For z, it varies from 0 to the upper boundary defined by the cone equation
`\( z = √(x^2 + y^2) \),` which in cylindrical coordinates is simply
`\( z = r \).`

Therefore, the integral for the surface area S of the portion of the cone within the cylinder is:

`\[ S = \int_0^(2\pi) \int_0^6 r \, dz \, dr \, d\theta \]`

Solving this integral step by step:

`\[ S = \int_0^(2\pi) \int_0^6 r \, dz \, dr \, d\theta \]\[ = \int_0^(2\pi) \left[ z \right]_0^6 \, dr \, d\theta \]\[ = \int_0^(2\pi) 6r \, dr \, d\theta \]\[ = \left[ 3r^2 \right]_0^6 \, d\theta \]\[ = 3(6)^2 \int_0^(2\pi) d\theta \]\[ = 3(36)(2\pi) \]\[ = 216\pi \]`