Consider the portion of the Cartesian plan delimited, in the first quadrant, from $x=0$, from $y=0$ and from the circumference of radius $= 1$ with center in the point $(0; 1)$, and determine the volume of the solid that generates a full rotation around $x=0$.
Progress
This is my tentative of solving the problem but i don't understand why the book says that the result is $\pi/ 3$.
$x^2 + (y-1)^2 = 1$ is the equation of the circumference centered on $(0; 1)$
$x^2 + y^2 + 1 - 2y = 1$
$x^2 + y^2 -2y = 0$
$x = \sqrt{2y - y^2}$
I don't understand what surface I have to rotate.
Could you help me, please?
I don't understand if the surface to rotate is the pseudotriangle 0AB (in red), or the semicircle coloured in green.
Of course the center of the circle is in $(0; 1)$, the length of 0A is 1, and the length of AB is 1.
Thinking about this problem, I realize that - in my opinion - the green surface measures:
PiGreco * 1^2 /2 = PiGreco / 2
A full rotation around x=0 is a sphere. The volume of a sphere is: 4 PiGreco * R^3 / 3
We have R = 1 so the volums SEEMS to be: 4 PiGreco / 3.
BUT the book says that the result is PiGreco / 3
Who is right? Who is wrong? Did I misunderstood the surface to rotate around x=0?
As you found, the volume of the green part is $4\pi/3$, so it can't be the one they want.
The volume of the red part can be found geometrically, without integration. Think of rotating the vertical line segment with endpoints $(1;0)$ and $(1;1)$ around the axis $x=0$. This creates a cylinder of height $2$ and radius $1$, hence of volume $2\pi$. Subtract the volume of the sphere which you know, and you'll get the red part together with its mirror image near the top of cylinder. Divide by two to get the volume of the red part.
Aside: Archimedes found that the volume of a ball inscribed into a cylinder is $2/3$ of the volume of the cylinder. He was so proud of this result that he asked for a sketch of a sphere inscribed in a cylinder to be on his grave. On the Sphere and Cylinder