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junior college 2 | H2 Maths
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Date Posted: 4 years ago
Views: 630
Find the point of intersection first.
x³ = √x
x^6 = x
x^6 - x = 0
x(x^5 - 1) = 0
x = 0 or x^5 = 1
x = 0 or x = 1^(1/5) = 1
x³ = √x
x^6 = x
x^6 - x = 0
x(x^5 - 1) = 0
x = 0 or x^5 = 1
x = 0 or x = 1^(1/5) = 1
Area of R
= ∫¹0 (√x - x³) dx
= ∫¹0 (x^½ - x³) dx
= [ ⅔ x³/² - ¼x⁴]¹0
= ⅔(1)³/² - ¼(1)⁴
= ⅔ - ¼
= 5/12
= ∫¹0 (√x - x³) dx
= ∫¹0 (x^½ - x³) dx
= [ ⅔ x³/² - ¼x⁴]¹0
= ⅔(1)³/² - ¼(1)⁴
= ⅔ - ¼
= 5/12
Volume of revolution around y axis :
For y = √x
y² = x
x² = y⁴
For y = x³
x = y^⅓
x² = y^⅔
when x = 1, y = 1³ = 1
So the point of intersection is (1,1)
volume of revolution : π ∫ x² dy
= π ∫¹0 y^⅔ dy - π ∫¹0 y⁴ dy
= π ∫¹0 (y^⅔ - y⁴) dy
= π [3/5 y^(5/3) - 1/5 y^5]¹0
= π ( 3/5 (1)^(5/3) - 1/5 (1)^5 )
= π (3/5 - 1/5)
= 2/5 π
For y = √x
y² = x
x² = y⁴
For y = x³
x = y^⅓
x² = y^⅔
when x = 1, y = 1³ = 1
So the point of intersection is (1,1)
volume of revolution : π ∫ x² dy
= π ∫¹0 y^⅔ dy - π ∫¹0 y⁴ dy
= π ∫¹0 (y^⅔ - y⁴) dy
= π [3/5 y^(5/3) - 1/5 y^5]¹0
= π ( 3/5 (1)^(5/3) - 1/5 (1)^5 )
= π (3/5 - 1/5)
= 2/5 π
See 2 Answers
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Part one.
Date Posted:
4 years ago
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Part two.
Date Posted:
4 years ago
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