Asked by lizzy
                the region in the first quadrant enclosed by the curves y=0 x=2 x=5 and y=1/root(1+x) is rotated about the x axis. the volume of the solid generated is:
a. pie In(5)
b.pie In(3/4)
c.pie In(4/3)
d.pie In(10)
e. pie In(2)
            
        a. pie In(5)
b.pie In(3/4)
c.pie In(4/3)
d.pie In(10)
e. pie In(2)
Answers
                    Answered by
            mathhelper
            
    looks like you want
∫ 1/√(x+1) dx from 0 to 2
= ∫ (x+1)^(-1/2) dx from 0 to 2
= [ 2(x+1)^(1/2) ] from 0 to 2
= [2√(x+1) ] from 0 to 2
= 2√3 - 2√1
= 2√3 - 2
    
∫ 1/√(x+1) dx from 0 to 2
= ∫ (x+1)^(-1/2) dx from 0 to 2
= [ 2(x+1)^(1/2) ] from 0 to 2
= [2√(x+1) ] from 0 to 2
= 2√3 - 2√1
= 2√3 - 2
                    Answered by
            correction - mathhelper
            
    sorry, I found the area by mistake, you wanted the volume
vol = ∫ π y^2 dx from 0 to 2
= π ∫ 1/(x+1) dx from 0 to 2
= π [ ln(x1) ] from 0 to 2
= π(ln3 - ln1]
= π ln3
    
vol = ∫ π y^2 dx from 0 to 2
= π ∫ 1/(x+1) dx from 0 to 2
= π [ ln(x1) ] from 0 to 2
= π(ln3 - ln1]
= π ln3
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