Volume Under a Plane Find the volume under $z = 3x$ and about he $1$st quadrant area in the $xy$ plane bounded by
$$
x = 0,\quad y = 0,\quad x = 4,
\quad x^{2} + y^{2}=25,
$$
$$\int_{0}^{4}\int_{0}^{25-x^{2}}3x\,{\rm d}x\,{\rm d}y = 98.$$
Not sure if this is correct. I was confused if this took the volume of the area b/w $$4<x<5 $$ and $$0<y<\sqrt{25-x^{2}}$$ as well, so a second integral would be needed to remove this portion of volume.
I appreciate the help. Thanks

$$Volume= \int_{0}^4 \int_{0}^{\sqrt{25-x^2}} 3x \ dy \ dx $$
