Answer:
The answer is: B
Step-by-step explanation:
Hope this helps!
The curve

is parameterized by

so in the line integral, we have





You are mistaken in thinking that the gradient theorem applies here. Recall that for a scalar function

, we have gradient

. The theorem itself then says that the line integral of

along a curve

parameterized by

, where

, is given by

Specifically, in order for this theorem to even be considered in the first place, we would need to be integrating with respect to a vector field.
But this isn't the case: we're integrating

, a scalar function.
Answer: h=3v/pir^2 +1
Step-by-step explanation:
sorry i can't use the pi symbol so i'll be referring to pi as pi
^2 is squared
1/3pir^2(h-1) =v
<u>simplify 1/3pir^2</u>
pi/3r^2(h-1)=v
<u>multiply both sides by 3 and simplify </u>
3*pi/3r^2(h-1)=3v <-- simplify = pir^2(h-1)=3v
<u>divide both sides by pir^2</u>
pir^2(h-1)/ pir^2 = 3v/ pir^2
<u>simplify </u>
h-1 = 3v/pir^2
<u>add 1 to both sides</u>
h-1 +1 = 3v/pir^2+1
answer/ simplify:
h=3v/pir^2 +1