To find volume of solid with $x=y^2$ and $x=9$ perpendicular to $x$-axis and cross-section to be taken is triangle with $h=\frac b4$.
I am confused in imagining the triangle.
$$\text{Volume to me} = \int_0^9\left(\frac12 \cdot x \cdot \frac x4\right)\,\mathrm dx $$
Am I correct?
The base of the triangle should be line segment perpendicular to x axis and confined in the curve. So the length of the base is $2\sqrt {x} $. Then the area of the cross sectional triangle is $\frac {1}{2} \cdot 2\sqrt {x} \frac {1}{4} \cdot 2\sqrt {x} =\frac {1}{2}x$. And that is the integrand. The limits are correct.
Below I sketched the 2D base and 3D picture of them: