Question

5. Show that the surface area of the solid region bounded by the three cylinders x2 + y2 = 1, y2 +z2 = 1 and x2 +z2 = 1 is 48 – 24V2. + =

Answer 1

To find the surface area of the solid region bounded by the three cylinders x^2 + y^2 = 1, y^2 + z^2 = 1, and x^2 + z^2 = 1, we can use the divergence theorem to convert the surface integral to a volume integral.

The surface integral of the vector field F = xi + yj + zk over the boundary of the solid region is:

∬_S F . dS

where S is the boundary of the solid region. By the divergence theorem, this is equal to:

∭_V div(F) dV

where V is the solid region enclosed by S, and div(F) is the divergence of F.

The divergence of F is:

div(F) = ∂/∂x (x) + ∂/∂y (y) + ∂/∂z (z) = 3

Since the divergence is constant, we can pull it out of the integral and integrate over the volume of the solid region. The limits of integration are:

-1 ≤ x ≤ 1
-√(1 - x^2) ≤ y ≤ √(1 - x^2)
-√(1 - x^2 - y^2) ≤ z ≤ √(1 - x^2 - y^2)

The volume integral is:

∭_V 3 dV = 3V

where V is the volume of the solid region.

To find the volume of the solid region, we can integrate over the cylindrical coordinates:

V = ∫_0^1 ∫_0^2π ∫_0^(√(1 - r^2)) r dz dθ dr

= ∫_0^1 ∫_0^2π r(√(1 - r^2)) dθ dr

= ∫_0^1 -cos(θ)|_0^2π dr

= 2

Therefore, the surface area of the solid region is:

∭_V div(F) dV = 3V = 6

So the surface area of the solid region bounded by the three cylinders x^2 + y^2 = 1, y^2 + z^2 = 1, and x^2 + z^2 = 1 is 6.

Answer 2

To find the surface area of the solid region bounded by the three cylinders x^2 + y^2 = 1, y^2 + z^2 = 1, and x^2 + z^2 = 1, we need to calculate the surface areas of the shared regions between the cylinders. By using the formulas for the surface area of cylinders and circles, we can find the surface area of each shared region and add them up to get the total surface area of the solid region. The total surface area is approximately 15.71.

Step-by-step explanation:

The three cylinders are:

x2 + y2 = 1 (cylinder 1)

y2 + z2 = 1 (cylinder 2)

x2 + z2 = 1 (cylinder 3)

To find the surface area of the solid region bounded by these cylinders, we need to consider the shared regions.

The shared region 1 is the intersection of cylinder 1 and cylinder 2, which lies in the yz-plane. Its surface area can be calculated using the formula for the surface area of a cylinder with radius r and height h: 2πrh.

The shared region 2 is the intersection of cylinder 2 and cylinder 3, which lies in the xz-plane. Its surface area can also be calculated using the formula for the surface area of a cylinder: 2πrh.

The shared region 3 is the intersection of cylinder 1 and cylinder 3, which lies in the xy-plane. Its surface area can be calculated using the formula for the surface area of a circle with radius r: πr2.

To find the total surface area of the solid region, we need to calculate the surface areas of the three shared regions and add them up:

Total Surface Area = Surface Area of Shared Region 1 + Surface Area of Shared Region 2 + Surface Area of Shared Region 3

Substituting the appropriate values, we get:

Total Surface Area = 2π(r1)(h1) + 2π(r2)(h2) + π(r3)2

Since all three cylinders have a radius of 1 and height 1, we can simplify the equation:

Total Surface Area = 2π(1)(1) + 2π(1)(1) + π(1)2

Total Surface Area = 2π + 2π + π = 5π

Using the value of π as approximately 3.1416, we can calculate the total surface area:

Total Surface Area ≈ 5 × 3.1416 ≈ 15.708

Rounding to two decimal places, the surface area of the solid region bounded by the three cylinders is approximately 15.71.