Turning points question

Anthonyk2013

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Sep 15, 2013
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Wondering what's going on in this part of the solution.
 

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They took the 1/2 outside of the integration, and integrated the expression that was left, i.e (y+3)

y+3, when integrated w.r.t dy, becomes (y^2/2)+3y
 
They took the 1/2 outside of the integration, and integrated the expression that was left, i.e (y+3)

y+3, when integrated w.r.t dy, becomes (y^2/2)+3y


Thanks i I see it now. But what happened to the x^2 on the above line.
 
Nothing happened to it. The solution in the image you posted derives a value for x2 in terms of y:

\(\displaystyle x^2=\frac{y-3}{2}\)

Then take that value and plug it into the integral:

\(\displaystyle \displaystyle \text{Volume}_{y-axis}=\int _3^{21}\:\pi \:x^2dy=\pi \int _3^{21}\:\frac{y-3}{2}dy\)
 
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