S = ∫[-13,3] 2πx√(1+x'^2) dy
x = √(3-y)
x' = -1/2√(3-y)
1+x'^2 = 1 + 1/4(3-y) = (13-4y)/(3-y)
So,
S = ∫[-13,3] 2π√(3-y)(13-4y)/(3-y) dy
= 2π∫[-13,3] (13-4y)/√(3-y) dy
= 2π (2/3 (4y-15)√(3-y)) [-13,3]
= 1072/3 π
The given curve is rotated about the y-axis. Find the area of the resulting surface.
y = 3 − x2, 0 ¡Ü x ¡Ü 4
4 answers
it says that is wrong
so, what headway have you made? See any mistakes in my calculations?
I redid the problem using polar coordinates, where I set the (0,0,0) at the base of the paraboloid section, so that
z = 3 - (x^2+y^2)
That means that we have only to evaluate
S = ∫[0,2π]∫[0,√3] r√(1+4r^2) dr dθ
and we get π/6 (13√13 - 1) = 24.0186π
Not sure where I went wrong using x = g(y) and rectangular coordinates. You can see a discussion of the coordinate change at
http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/surface/surface.html
z = 3 - (x^2+y^2)
That means that we have only to evaluate
S = ∫[0,2π]∫[0,√3] r√(1+4r^2) dr dθ
and we get π/6 (13√13 - 1) = 24.0186π
Not sure where I went wrong using x = g(y) and rectangular coordinates. You can see a discussion of the coordinate change at
http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/surface/surface.html
1 answer