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
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
I am just suppose to find the surface area. We are studying arc length
To find the exact area of the surface obtained by rotating the curve about the x-axis, we can use the disk method.
First, let's find the equation of the curve in terms of y. Rearranging the given equation, we have:
y^2 = 9x - 36
Now, we need to solve this equation for x to express the curve in terms of y:
x = (y^2 + 36) / 9
Next, we integrate the equation of the curve from y = √36 = 6 to y = √(9(8) - 36) = √12 = 2√3. This will give us the area of one half of the surface. Since the curve is symmetric about the x-axis, we can double this area to get the total surface area.
Therefore, the integral we need to evaluate is:
A = 2π ∫[6, 2√3] ( (y^2 + 36) / 9 ) * √(1 + (dy/dx)^2 ) dy
To find (dy/dx), we differentiate x = (y^2 + 36) / 9 with respect to y:
dx/dy = (2y) / 9
Now, we can substitute this value into the expression for √(1 + (dy/dx)^2):
√(1 + (dy/dx)^2 ) = √(1 + (2y/9)^2) = √(1 + 4y^2/81)
Substituting this back into the integral, we have:
A = 2π ∫[6,2√3] ( (y^2 + 36) / 9 ) √(1 + 4y^2/81) dy
Evaluating this integral will give us the exact area of the surface obtained by rotating the curve about the x-axis.
Please note that the integral involves some complicated algebraic manipulations and may not have a simple closed-form solution. In this case, you may need to rely on numerical methods or approximate the value of the integral using numerical approximation techniques.
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.
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