Solution:
As region bounded by y-axis, the line y=6, and the line y=1/2 is a line segment of definite length on y-axis.
We consider a line , one dimensional if it's thickness is negligible.
So, Line is two dimensional if it's thickness is not negligible becomes a quadrilateral.
So, Area (region bounded by y-axis, the line y=6, and the line y=1/2 is a line segment of definite length on y-axis)= Area of line segment between [,y=6 and y=1/2.]= 6-1/2=11/2 units if we consider thickness of line as negligible.
Here’s the answer and working out. Hope it helps
Answer:
Step-by-step explanation:
In general (x, y, z) = (p ·cos s ·sin t, p· sin s· sin t, p · cos t) where p is radius
x(s, t)= 3 ·cos s ·sin t
y(s, t)= 3· sin s· sin t
z(s, t) = 3· cos t
Where, radius is 3, s ∈ [0,2π), and t ∈ [0, π]