A giant soda glass is formed by rotating the graph of
f(x)= .03x^2+1.125, 0<=x<5
f(x)= -.2(x-5)^2+1.875, 5<=x<6
f(x).1(x-6)^2+1.675, 6<=x<=7
about the x axis
a) If the measurements are in inches, find the capacity of the glass in cubic inches
B) If 16.3787 cm^3=1in^3 and 1000cm^3=1 liter, will the glass hold more than .5 liters of soda? Justify your conclusion
I tried graphing the piecewise function and I did not understand how it could be revolved around the x axis or relate to this problem
To find the capacity of the glass in cubic inches, we can use the method of cylindrical shells to calculate the volume.
First, let's visualize the graph of the function and understand how it forms the shape of the glass. The given function is a piecewise function defined on different intervals. The graph of each interval represents a part of the glass.
On the interval 0 ≤ x < 5:
The function is f(x) = 0.03x^2 + 1.125, which represents the top portion of the glass. This is a quadratic function that opens upward.
On the interval 5 ≤ x < 6:
The function is f(x) = -0.2(x-5)^2 + 1.875, which represents the middle portion of the glass. This is also a quadratic function that opens downward, giving the shape of a concave lens.
On the interval 6 ≤ x ≤ 7:
The function is f(x) = 0.1(x-6)^2 + 1.675, which represents the bottom portion of the glass. This is another quadratic function that opens upward, similar to the top portion.
To rotate this graph about the x-axis and form the glass, we imagine each vertical cross-section of the graph rotating 360 degrees to create a cylindrical shell.
Now let's calculate the volume of each cylindrical shell, which represents an infinitesimally thin slice of the glass. The formula for the volume of a cylindrical shell is V = 2πrh * ∆x, where r represents the radius and h represents the height of the shell. We need to calculate the sum of all these volumes for each interval.
On the interval 0 ≤ x < 5:
Since the function is on top of the x-axis, the height of the shell is f(x). The radius is x, as it ranges from 0 to x. Hence, the volume of each shell is V1 = 2πxf(x) * ∆x. We need to integrate this expression from 0 to 5 to cover the entire range of this interval.
On the interval 5 ≤ x < 6:
The height of the shell is still f(x), but since the graph is below the x-axis, we need to consider the absolute value of f(x). The radius of the shell is 5 - x, as x ranges from 5 to x. Hence, the volume of each shell is V2 = 2π(5-x)|f(x)| * ∆x. We need to integrate this expression from 5 to 6.
On the interval 6 ≤ x ≤ 7:
The height of the shell is still f(x), and since the graph is again above the x-axis, the radius is x - 6. Hence, the volume of each shell is V3 = 2π(x-6)f(x) * ∆x. We need to integrate this expression from 6 to 7.
To find the total volume of the glass, we need to sum up all these volumes:
V_total = ∫[0,5] V1 + ∫[5,6] V2 + ∫[6,7] V3
By evaluating this integral, you can find the capacity of the glass in cubic inches.
Regarding part B, we know that 16.3787 cm^3 = 1 in^3 and 1000 cm^3 = 1 liter. So, we need to convert the capacity of the glass from cubic inches to liters.
Let's assume the capacity of the glass in cubic inches is V_glass_in^3. We can convert it to liters by using the conversion factor:
V_glass_liters = V_glass_in^3 / (16.3787 cm^3/in^3 * 1000 cm^3/liter)
If the resulting V_glass_liters is greater than 0.5 liters, then the glass can hold more than 0.5 liters of soda.
By following these steps, you should be able to find the capacity of the glass and determine whether it can hold more than 0.5 liters of soda or not.