Find the area of the surface generated by revolving the arc
about the x-axis
I'm not sure how to graph or where to begin
No need to graph: pay attention that y = t^2/2 = t^[(2/3)]^3/2 = x^3/2, so you need the integral INT{0,2} (x^3/2)^2 dx and then multiply this by Pi = 3.14159...and that's the area.
No need to graph: pay attention that y = t^2/2 = t^[(2/3)]^3/2 = x^3/2, so you need the integral INT{0,2} (x^3/2)^2 dx and then multiply this by Pi = 3.14159...and that's the area.
Uuh?? We're given y = t^2/2 , but we already know t^2 = x^3 ==>
y = x^3/2.
I think you should read more carefully the answers you're given and give a little more thinking time to them before rushing to ask back stuff.
No need to graph: pay attention that y = t^2/2 = t^[(2/3)]^3/2 = x^3/2, so you need the integral INT{0,2} (x^3/2)^2 dx and then multiply this by Pi = 3.14159...and that's the area.
Tonio
so i keep trying this and i get 14.36 instead of the correct answer of 13.676
so i keep trying this and i get 14.36 instead of the correct answer of 13.676
this is what i am doing:
If x = t^(2/3) and 0 <= t <= 2, then 0 <= x <= 4^(1/3) and not 2 ==> the integral's upper limit is 4^(1/3). But any way you don't get what you say the answer is.