Question
Find the exact area of the surface obtained by rotating the curve about the x-axis.
9x = y2 + 36, 4 ≤ x ≤ 8
I keep on getting 4pi (3^(3/2) +1)
so I tried to put the answer in that way and also 24.78pi
But it is wrong
9x = y2 + 36, 4 ≤ x ≤ 8
I keep on getting 4pi (3^(3/2) +1)
so I tried to put the answer in that way and also 24.78pi
But it is wrong
Answers
Steve
9x = y^2 + 36
I assume you just mean the part above the x-axis, since the curve is symmetric about that axis. So, using discs, the volume is
v = ∫[4,8] πr^2 dx
where r^2 = y^2 = 9x-36
v = π∫[4,8] 9x-36 dx
= π(9/2 x^2 - 36x) [4,8]
= 72π
Using shells, we have
v = ∫[0,6] 2πrh dy
where r = y and h = (8-x) = 8-(y^2+36)/9)
v = 2π∫[0,6] y(8-(y^2+36)/9) dy
2π(2y^2 - y^4/36) [0,6]
= 72π
I assume you just mean the part above the x-axis, since the curve is symmetric about that axis. So, using discs, the volume is
v = ∫[4,8] πr^2 dx
where r^2 = y^2 = 9x-36
v = π∫[4,8] 9x-36 dx
= π(9/2 x^2 - 36x) [4,8]
= 72π
Using shells, we have
v = ∫[0,6] 2πrh dy
where r = y and h = (8-x) = 8-(y^2+36)/9)
v = 2π∫[0,6] y(8-(y^2+36)/9) dy
2π(2y^2 - y^4/36) [0,6]
= 72π
sam
I am just suppose to find the surface area. We are studying arc length
Steve
Oops. Oh well. The observation about symmetry still holds. The surface area is just the sum of circumferences of many circles of radius y. So,
dS = 2πy ds
S = 2π∫[4,8] y√(1+y'^2) dx
9x = y^2+36, so y = 3√(x-4)
18 = 2yy'
y' = 9/y = 9/√(9x-36)
so, y'^2 = 81/(9x-36) = 9/(x-4)
S = 2π∫[4,8] (3√(x-4))√((x+5)/(x-4)) dx
= 6π∫[4,8] √(x+5) dx
= 4π(x+5)^(3/2) [4,8]
= 79.49 π
Better double-check my math.
dS = 2πy ds
S = 2π∫[4,8] y√(1+y'^2) dx
9x = y^2+36, so y = 3√(x-4)
18 = 2yy'
y' = 9/y = 9/√(9x-36)
so, y'^2 = 81/(9x-36) = 9/(x-4)
S = 2π∫[4,8] (3√(x-4))√((x+5)/(x-4)) dx
= 6π∫[4,8] √(x+5) dx
= 4π(x+5)^(3/2) [4,8]
= 79.49 π
Better double-check my math.
Steve
I redid the problem using polar coordinates, where I set the (0,0,0) at the base of the paraboloid section, so that
z = 4 - (x^2+y^2)/9
That means that we have only to evaluate
S = 1/9 ∫[0,2π]∫[0,6] r√(81+4r^2) dr dθ
and we get a nice tidy 49π
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 = 4 - (x^2+y^2)/9
That means that we have only to evaluate
S = 1/9 ∫[0,2π]∫[0,6] r√(81+4r^2) dr dθ
and we get a nice tidy 49π
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