Answer: The graph of this function V(r) is in the I (first) quadrant.
Solution:
V(r)=5(3.14)r^2
V(r)=15.7r^2
The graph of this function is a parabola with vertex at the origin and opens upward (I and II quadrants), but the radius is a positive number (right to the y-axis), then the graph of this function is in the I (first) quadrant.