Solve Volume of Solid: y=x, y=x^1/2, y=1

[necessary]

volume of solid:(

hi..I tried to solve it but ı couldn't .book says the answer is pi /6...please help me.

question is; y=x and y=x^1/2 about y =1

 

[arildno]

Pinpoint your problem.

 

[necessary]

ok I showed my work..


I used this formula
integral Pı(outer r)^2-Pı(inner)^2



inside integral from y=0 to y=1 pi(1-y)^2 - pi(1-y^2)^2

my result is 7/6 pi..:( where am I wrong??

 

[courtrigrad]

integrate it with respect to x. The inner radius is $$1-\sqrt{x}$$ and the outer radius is $$1- x$$.

So you have $$\int_{0}^{1} \pi((1-x)^{2}-(1-\sqrt{x})^{2}) \; dx$$

 

[arildno]

Well, your integrand is utterly wrong!
You are to rotate the VERTICAL strip $\sqrt{x}\leq{y}\leq{x}$ around y=1.
That gives you the integral:
$$V=\int_{0}^{1}\pi((1-x)^{2}-(1-\sqrt{x})^{2})dx=\pi\int_{0}^{1}(x^{2}-3x+2\sqrt{x})dx$$

 

[arildno]

Apart from being wrongly signed, you have rotated your figure around the line x=1, rather than around y=1.
That accounts for your discrepancy.
(Note, for example, that your INNER radius is the actual OUTER radius)