All works must be seen!

Find the volume of a solid torus which is obtained by rotating the circle $(x-R)^2+y^2=r^2$ with $R>r>0$ about the $y$-axis.

Let $C$ be a curve parametrized by $(x(t),y(t))$. We assume:

  1. the curve intersects $y$-axis at $t=a$ and $t=b$ with $y(a)<y(b)$;
  2. the curve intersects with $y=c$ at exactly one point for all $c$ between $y(a)$ and $y(b)$;
  3. $x(t)>0$ for all $t\in (a,b)$. Explain that the area inclosed by $C$ and $y$-axis is \(\left|\int_a^b x(t)y'(t)dt\right|.\) (I want to see how you partition the entire area into tiny rectangles, and use $x(t)$ and $y(t)$ to express the area of the rectangles.)

Suppose $f(x)$ is a nonegative and continuous function on $[a,b]$. Find the area of the surface revolving the curve about $x$-axis, and explain your formula.

Consider a segment of the curve described by the equation $x^{2/3}+y^{2/3}=1$ in the first quaudrant.

  1. Find the arc length of the curve.
  2. Find the area of the surface revolving the curve about $x$-axis.