I want to calculate the volume of the solid of revolution around the x-axis of this figure
$x = (1-t^2)/(t^4+4)$
$y = (t+1)*(1-t^2)/(t^4+1)$
for t between -1 and 1. In the figure below the plot is shown
In my opinion to do this I have to calculate the integral of
$\int_{0}^{0.25}\pi * (y_{upper}^2 – y_{lower}^2) dx$.
The solution in my book is
$\int_{-1}^{1}\pi * y(t)^2 * (dx/dt) \text{dt} $
I do not understand why after changing the variable of integration to $t$ the upper and lower part of the curve or not subtracted anymore.
Divide the integral into two parts, one from $0$ to $1$ and one from $-1$ to $0$. In fact you should plot $(x(t),y(t))$ separately for those two intervals. You will notice that these two intervals form the upper and lower branches of your curve. Moreover, when you vary $t$ from $-1$ you go on the lower curve from $(0,0)$ and return on the upper branch. But that means that you increase $x$ then you decrease it. You can therefore write $$\int_0^{0.25}(y_{upper}^2-y_{lower}^2)dx=\int_0^{0.25}y_{upper}^2dx-\int_0^{0.25}y_{lower}^2dx=\int_0^{0.25}y_{upper}^2dx+\int_{0.25}^0y_{lower}^2dx$$