A glass tunnel in the shape of a half cylinder with length 58 feet and radius 7 feet is built along the bottom of an aquarium that is 43 feet deep. Set up an integral that would find the fluid force on the tunnel (water density 62.5 lbs/ft³) 1.₁ -7 Force = dx
A glass tunnel in the shape of a half cylinder with length 58 feet and radius 7 feet is built along the bottom of an aquarium that is 43 feet deep. Set up an integral that would find the fluid force on the tunnel (water density 62.5 lbs/ft³) 1.₁ -7 Force = dx
Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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![**Fluid Force on a Glass Tunnel in an Aquarium**
*Problem Statement:*
A glass tunnel in the shape of a half-cylinder with a length of 58 feet and a radius of 7 feet is constructed along the bottom of an aquarium that is 43 feet deep. Given that the water density is 62.5 lbs/ft³, set up an integral to calculate the fluid force exerted on the tunnel.
*Integral Representation:*
To find the fluid force on the tunnel, we need to set up an integral that accounts for the varying pressure exerted by the water at different depths. The force in this context can be expressed using the following integral:
\[ \text{Force} = \int_{-7}^{7} \quad dx \]
**Explanation of Variables and Integral:**
- The integral limits \(-7\) to \(7\) represent the radius of the half-cylinder in feet.
- The variable \(dx\) indicates a differential element along the horizontal distance across the tunnel's radius.
The actual complete formulation would require including the water pressure formula \(P = \rho gh\), where:
- \(\rho\) is the water density (62.5 lbs/ft³),
- \(g\) is the acceleration due to gravity (often approximated as 32.2 ft/s²),
- \(h\) is the depth of water above a specific point, which could be related to the radius and specific coordinate of the half-cylinder cross-section being integrated.
However, the provided image does not include the complete integrand of this integral for fluid force calculation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb4c907f3-6692-4a58-8d85-7b416968270c%2Fcbcf039f-11b5-46e9-8edf-056db9023fc7%2F64btq5f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Fluid Force on a Glass Tunnel in an Aquarium**
*Problem Statement:*
A glass tunnel in the shape of a half-cylinder with a length of 58 feet and a radius of 7 feet is constructed along the bottom of an aquarium that is 43 feet deep. Given that the water density is 62.5 lbs/ft³, set up an integral to calculate the fluid force exerted on the tunnel.
*Integral Representation:*
To find the fluid force on the tunnel, we need to set up an integral that accounts for the varying pressure exerted by the water at different depths. The force in this context can be expressed using the following integral:
\[ \text{Force} = \int_{-7}^{7} \quad dx \]
**Explanation of Variables and Integral:**
- The integral limits \(-7\) to \(7\) represent the radius of the half-cylinder in feet.
- The variable \(dx\) indicates a differential element along the horizontal distance across the tunnel's radius.
The actual complete formulation would require including the water pressure formula \(P = \rho gh\), where:
- \(\rho\) is the water density (62.5 lbs/ft³),
- \(g\) is the acceleration due to gravity (often approximated as 32.2 ft/s²),
- \(h\) is the depth of water above a specific point, which could be related to the radius and specific coordinate of the half-cylinder cross-section being integrated.
However, the provided image does not include the complete integrand of this integral for fluid force calculation.
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