How do you do this?
Find the volume generated by revolving the area bounded by the ellipse(x^2)/9 + (y^2)/4 = 1 about the line x=3.
Thank you Mr. Steve! :3
Well, calculating the volume of a solid of revolution can be quite elliptic! But fear not, I promise not to go off on a tangent!
Since we are revolving the ellipse around the vertical line x = 3, we can imagine the resulting shape as a three-dimensional 'lens' with a skinny waist in the middle. Let's call it an "elliptical lemonade glass" for lack of a better term!
To find the volume of this 'lemonade glass,' we need to integrate the cross-sectional areas from its top (the rounded part) to its bottom (the two flat bases).
First, let's express the equation of the ellipse as y = f(x). Rearranging the equation, we have:
(y^2)/4 = 1 - (x^2)/9
y^2 = 4 - (4/9)x^2
y = ±sqrt(4 - (4/9)x^2)
Now, using the symmetry of the ellipse, we can focus on just one side and then double our result later! Considering the positive side only, the distance between the ellipse and the line x = 3 is given by r = 3 - x.
To find the cross-sectional area at any given x, we'll use the formula for the area of a circle with radius r:
A = π*r^2 = π*(3 - x)^2
Next, we integrate this equation with respect to x over the interval [-3, 3] (the range of x for the ellipse):
V = 2 * ∫[from -3 to 3] π*(3 - x)^2 dx
After evaluating this integral, you'll have the volume of our elliptical lemonade glass! Just be careful not to confuse it with a mathematical cocktail shaker! Cheers!
To find the volume generated by revolving the area bounded by the ellipse about the line x = 3, we can use the method of cylindrical shells.
First, let's graph the given ellipse. The equation of the ellipse is (x^2)/9 + (y^2)/4 = 1. This is an ellipse centered at the origin with a major axis of length 6 (2a = 6) and a minor axis of length 4 (2b = 4).
Next, we need to determine the limits of integration for the volume integral. Since we are revolving the ellipse about the line x = 3, the axis of revolution is at x = 3. The interval of integration for x is from 3 - a to 3 + a, where a represents the length of the semi-major axis.
Since the semi-major axis is 3, the limits of integration for x are from 0 to 6.
Now, let's set up the integral for the volume using the method of cylindrical shells. The volume generated by each cylindrical shell is given by the formula V = 2πrhΔx, where r is the radius of the shell, h is the height of the shell, and Δx is the width of the shell.
In this case, the radius r of each shell is given by r = y, the height h of each shell is given by h = 2x, and Δx is just dx. So the volume of each shell becomes V = 2πy(2x)dx.
To express y in terms of x, we rearrange the equation of the ellipse: (x^2)/9 + (y^2)/4 = 1. Solving for y, we get y = 2√(1 - (x^2)/9).
Substituting y = 2√(1 - (x^2)/9) into the volume integral equation, we have V = 2π(2√(1 - (x^2)/9))(2x)dx.
We can simplify this to V = 8π√(1 - (x^2)/9)x dx.
Now, we can integrate this equation from x = 0 to x = 6 to find the volume: V = ∫[0 to 6] 8π√(1 - (x^2)/9)x dx.
To solve this integral, you can use techniques such as substitution or integration by parts.
we have
4x^2 + 9y^2 = 36
The line x=3 lies at one end of the ellipse, so shells are probably the way to go. The volume v is thus
v = ∫[-3,3] 2πrh dx where
r = 3-x and h = y = √(36-4x^2)/3
v = 2π/3 ∫[-3,3] (3-x)√(36-4x^2) dx
Now just break it into two pieces, with one a straightforward u substitution, and the other a trig substitution.
with thanks to wolframalpha.com, we end up with
v = 18π^2