The given curve is rotated about the y-axis. Find the area of the resulting surface.
y = 3 − x2, 0 ¡Ü x ¡Ü 4
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
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
To find the area of the resulting surface, we can use the surface area formula for revolution about the y-axis.
The formula is given by:
S = 2π ∫ [x1, x2] y(x) √(1 + (dy/dx)²) dx
In this case, we're given the equation y = 3 - x^2 and the limits of integration are 0 ≤ x ≤ 4.
First, let's find dy/dx:
dy/dx = d(3 - x^2)/dx
= -2x
Now, let's find the limits of integration with respect to y.
When x = 0, y = 3 - 0^2 = 3.
When x = 4, y = 3 - 4^2 = -13.
So, the limits of integration with respect to y are 3 ≤ y ≤ -13.
Next, let's express x in terms of y:
y = 3 - x^2
x^2 = 3 - y
x = ±√(3 - y)
Since the curve is symmetric about the y-axis, we only need to consider the positive x-values.
Now we can rewrite the surface area formula as:
S = 2π ∫ [y1, y2] x(y) √(1 + (dx/dy)²) dy
S = 2π ∫ [3, -13] √(3 - y) √(1 + (1/(-2√(3-y)))²) dy
Simplifying the expression inside the integral:
S = 2π ∫ [3, -13] √(3 - y) √(1 - 1/(4(3-y))) dy
Now we can integrate to find the area:
S = 2π ∫ [3, -13] √(3 - y) √((4(3-y) - 1)/(4(3-y))) dy
= 2π ∫ [3, -13] √(12 - 4y - y²) / √(4(3-y)) dy
This is the integral we need to evaluate to find the area of the resulting surface.