The area below the curve above x = 0 is

In[4]:= Integrate[y - y^2, {y, 0, 1}]

Out[4]= 1/6

The look of the surface of revolution is given by

RevolutionPlot3D[y - y^2, {y, 0, 1}]

The get the volume remember what one does to get the volume of the torus: If R is the radius of rotation of the circle with radius r (the actual tube of the torus), then the volume of the torus is 2 Pi R Pi r^2 = 2 Pi^2 R r^2, R >= r. Rotating your curve, obviously R = 1/2 so the volume is 2 Pi 1/2 1/6 = Pi/6. Check this with WolframAlpha

http://www.wolframalpha.com/input/?i=volume+of+revolution

querying "Volumn of revolution" and using the given formula

In[6]:= 2 \[Pi] Integrate[y (y - y^2), {y, 0, 1}]

Out[6]= \[Pi]/6

In Wolfram Alpha one can not do that with exactly one query seemingly, but after one typed "volume of revolution" one sees the formula from In[6] above and then on might put it into WolframAlpha to get the integral there too, so one does it with two queries.

By the way, nearly every command from Mathematica might by typed into and answered by WolframAlpha, the restrictions are only

- the number of characters of the command
- the computational cost to get the answer

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