Respuesta :
Using the concepts of torus, we got that 241.264cm² is the change in change in surface area of the torus when r changes from r to r and r changes from r to r
We know very well that surface area of torus is given by 4π²(R²-r²) where R is the radius of the outer shell and r is the radius of the inner shell.
We are given that initial value of inner radius (r)=3.00,
final value of initial radius (r)=3.05.
initial value of outer radius (R)=5.50,
and final value of outer radius(R)=5.65
Therefore, according to the formula,
Initial surface area of the torus=4π²(R²-r²)
On putting the value of R and r,
=>S₁=4π²[(5.05)²-(3.00)²]
=>S₁=4×3.14×3.14×[25.5025-9]
=>S₁= 39.4384×16.5025
=>S₁=650.8321cm²
Similarly, for final surface area of torus
=>S₂=4π²[(5.65)²-(3.05)²]
=>S₂=4×3.14×3.14×[31.9225-9.3025]
=>S₂= 39.4384×22.62
=>S₂=892.09cm²
So, the change in surface area(ΔA)=S₂-S₁
=>ΔA=892.09-650.8321
=>ΔA=241.264cm²,
Hence, the change in surface area of the torus when r changes from r to r and r changes from r to r is 241.264cm²
To know more about torus, visit here:
https://brainly.com/question/14287542
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(Complete question):
The surface area of a torus with an inner radius r and an outer radius R>r is S = 4π²(R²-r²)². Estimate the change in the surface area of the torus when r changes from r=3.00cm to r=3.05cm and R changes from R=5.50cm to R=5.65cm
