Semi-circle cross section volume

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Homework Statement



base of a solid is bounded by y=\sqrt{x}, the x-axis, and x=9. each cross-section perpendicular to the x-axis is a semi-circle. find the volume of the solid

Homework Equations


The Attempt at a Solution


I found the answer to be 81\pi/16 by the following steps:
A(semicircle)=0.5pir^2=(pi/8)*d^2 (d=diameter), and d=y=sqrt(x), so A=(pi/8)*x
integrate (pi/8)*x from 0 to 9 (pi*9^2 /16), I get the answer as 81pi/16

BUT the answer provided by my book is 81pi/8, did I miss something? I'm so confused and please help.
 
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I agree with you. Maybe we are both missing something, but I can't think what it might be.
 
I agree with you both.
 
i guess the answer on my book is wrong then. I really can't think of any other ways
 
The book solution apparently calculated the volume assuming the cross section was a circle instead of a semi-circle.
 
LCKurtz said:
The book solution apparently calculated the volume assuming the cross section was a circle instead of a semi-circle.

ya, that's what I think too. The thing is, my book's never wrong before (for several times I doubted its answers and it turned out it's always correct), and this is the last question for the last unit. Hopefully it's just a decoy this time hehe.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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