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If the Area of the surface generated by revolving the curve y = x3, 0 ≤ x ≤ 1/2, about the x-axis is kπ/1728 then k is equal to .....
Correct answer is '61'. Can you explain this answer?
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If the Area of the surface generated by revolving the curve y = x3, 0&...
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If the Area of the surface generated by revolving the curve y = x3, 0&...
To find the area of the surface generated by revolving the curve y = x^3, 0 < x="" />< 1="" about="" the="" x-axis,="" we="" can="" use="" the="" formula="" for="" the="" surface="" area="" of="" a="" solid="" of="" />
The formula for the surface area of a solid of revolution is given by:
S = 2π ∫[a,b] y√(1+(dy/dx)^2) dx
In this case, the curve y = x^3 is being revolved around the x-axis, so our limits of integration are a = 0 and b = 1.
First, we need to find dy/dx:
dy/dx = 3x^2
Now, we can substitute this into the formula and integrate:
S = 2π ∫[0,1] x^3 √(1+(3x^2)^2) dx
Simplifying the expression under the square root:
S = 2π ∫[0,1] x^3 √(1+9x^4) dx
Now, we can integrate:
S = 2π ∫[0,1] x^3 √(1+9x^4) dx
= 2π/36 [(1+9x^4)^(3/2)] |[0,1]
= π/18 [(1+9)^(3/2) - 1]
Finally, we can simplify the expression:
S = π/18 [(10)^(3/2) - 1]
≈ π/18 (31.6227766 - 1)
≈ π/18 (30.6227766)
≈ (π/18) * 30.6227766
≈ 1.7015296
Therefore, the area of the surface generated by revolving the curve y = x^3, 0 < x="" />< 1="" about="" the="" x-axis="" is="" approximately="" 1.7015296="" units="" squared.="" 1="" about="" the="" x-axis="" is="" approximately="" 1.7015296="" units="" />
The formula for the surface area of a solid of revolution is given by:
S = 2π ∫[a,b] y√(1+(dy/dx)^2) dx
In this case, the curve y = x^3 is being revolved around the x-axis, so our limits of integration are a = 0 and b = 1.
First, we need to find dy/dx:
dy/dx = 3x^2
Now, we can substitute this into the formula and integrate:
S = 2π ∫[0,1] x^3 √(1+(3x^2)^2) dx
Simplifying the expression under the square root:
S = 2π ∫[0,1] x^3 √(1+9x^4) dx
Now, we can integrate:
S = 2π ∫[0,1] x^3 √(1+9x^4) dx
= 2π/36 [(1+9x^4)^(3/2)] |[0,1]
= π/18 [(1+9)^(3/2) - 1]
Finally, we can simplify the expression:
S = π/18 [(10)^(3/2) - 1]
≈ π/18 (31.6227766 - 1)
≈ π/18 (30.6227766)
≈ (π/18) * 30.6227766
≈ 1.7015296
Therefore, the area of the surface generated by revolving the curve y = x^3, 0 < x="" />< 1="" about="" the="" x-axis="" is="" approximately="" 1.7015296="" units="" squared.="" 1="" about="" the="" x-axis="" is="" approximately="" 1.7015296="" units="" />
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If the Area of the surface generated by revolving the curve y = x3, 0≤x≤1/2, about the x-axis is kπ/1728 then k is equal to .....Correct answer is '61'. Can you explain this answer?
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If the Area of the surface generated by revolving the curve y = x3, 0≤x≤1/2, about the x-axis is kπ/1728 then k is equal to .....Correct answer is '61'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about If the Area of the surface generated by revolving the curve y = x3, 0≤x≤1/2, about the x-axis is kπ/1728 then k is equal to .....Correct answer is '61'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the Area of the surface generated by revolving the curve y = x3, 0≤x≤1/2, about the x-axis is kπ/1728 then k is equal to .....Correct answer is '61'. Can you explain this answer?.
If the Area of the surface generated by revolving the curve y = x3, 0≤x≤1/2, about the x-axis is kπ/1728 then k is equal to .....Correct answer is '61'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about If the Area of the surface generated by revolving the curve y = x3, 0≤x≤1/2, about the x-axis is kπ/1728 then k is equal to .....Correct answer is '61'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the Area of the surface generated by revolving the curve y = x3, 0≤x≤1/2, about the x-axis is kπ/1728 then k is equal to .....Correct answer is '61'. Can you explain this answer?.
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