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Stabilo123

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Find the volume of the solid formed when y = (x+5)^2 is rotated about the y-axis between y=1 and y=4
 

Boonyak

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Not sure if im right but ill give it a go :O, we learn from mistakes i guess

[maths] A= \pi\int^{b}_{a} x^2dy[/maths]

[maths] y=(x+5)^2[/maths]
[maths] \sqrt y= \sqrt(x+5)^2[/maths]
[maths] = \sqrt y= x+5[/maths]
[maths] = y^\frac{1}{2}= x+5[/maths]
[maths] =y^\frac{1}{2}-5= x[/maths]
[maths] = (y^\frac{1}{2}-5)^2= x^2 [/maths]

[maths] A= \pi\int^{4}_{1} (y^\frac{1}{2}-5)^2dy[/maths]


Please correct if wrong
 
Last edited:

Aysce

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y^(1/2)= x+5
y^(1/2) - 5 = x

x^2 = [y^(1/2) -5]^2
 

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