Krdytkyu

2 0 $begingroup$ Compute volume of solid created by revolving the area bounded by $y=sqrt x$ , $y=0$ $x=1$ , $x=4$ , around the y axis. I understand that I can find the volume by integration of the $A(y)$ from $0$ to $2$ since these are the $y$ values of the intercepts of $x=1$ and $x=4$ . My understanding is that I would compute the area $(A(y) $ by: $pi$ $(R^2)-r^2)$ where $R$ is the outer radius and $r$ is the inner radius. Then I integrate this as $int_0^2 A(y)dy$ . So, I compute the outer radius $R$ by calculating the $x$ distance from the right most boundary, in this case $x=4$ and the $y$ axis. Thus $R=4$ . Is this the way to calculate $R$ even if part of the $R$ $x$ distance isn't inside the boundary?? Ugh,confused on the definition here. The inner radius is the $x$ distance f