Hello, Quasar!
Welcome aboard!
Your setup is almost correct . . .
Use the Shell Method to find the volume of the solid:
the region bounded by $\displaystyle y = 5x^{\frac{1}{2}},\:y=5,\:x=0$, about the line $\displaystyle x=1.$
Here is one of the ways I have tried setting this up.
After finding: $\displaystyle 5x^{\frac{1}{2}}= 5\quad\Rightarrow\quad x=1$
. . I Integrate on the xaxis from 0 to 1.
$\displaystyle V \;=\; 2\pi\int^1_0[\text{shell radius}][\text{shell height}]\,dx$ . . . Right!
. . $\displaystyle = \;2\pi\int^1_0(1x)\underbrace{(5x^{\frac{1}{2}})}_{\text{no}}\,dx$
Did you make a sketch? Code:

 :
5+     *
:::::* :
::* :
* :
 *     +      
0 1
The height is: $\displaystyle 5  5x^{\frac{1}{2}}$