Sketch the graphs of f(x) = x^2 and g(x) = 2x. Find a number a between 0 and 2 so that when the region between the graphs of f and g for 0 <= x <= a is revolved about the y-axis, the volume of the solid produced is 7pia3/12 . Please show all work
v = ∫[0,a] π(R^2-r^2) dx
= π∫[0,a] (2x)^2 - (x^2)^2 dx
= πa^3(4/3-a^2/5)
so, we want πa^3(4/3-a^2/5) = 7/12 π a^3
4/3 - a^2/5 = 7/12
80 - 12a^2 = 35
12a^2 = 55
a = √(55/12)
oops: √45/12 = √(15/2)
To sketch the graphs of f(x) = x^2 and g(x) = 2x, we can start by plotting some points and then connecting them to form smooth curves.
For f(x) = x^2, you can choose some values of x, calculate the corresponding y = f(x), and then plot the points. For example, when x = -2, f(-2) = (-2)^2 = 4, so you have the point (-2, 4). Repeat this process with other values of x, such as -1, 0, 1, and 2. Once you have enough points, connect them smoothly to form the graph of f(x) = x^2, which is a parabola opening upwards.
For g(x) = 2x, you can do the same process to plot some points and then connect them. For instance, when x = -2, g(-2) = 2(-2) = -4, so you have the point (-2, -4). Repeat this for other values of x, such as -1, 0, 1, and 2. Connect the points smoothly to form the graph of g(x) = 2x, which is a straight line passing through the origin.
Now let's find the number a between 0 and 2 such that the volume of the solid produced when the region between the graphs of f and g is revolved about the y-axis is 7πa^3/12.
To find this value of a, we need to set up an integral to calculate the volume of the solid of revolution. The volume of a solid of revolution can be calculated using the formula:
V = ∫[a,b] A(x) dx
Where V is the volume, A(x) is the cross-sectional area at a given value of x, and [a,b] is the interval over which we are integrating.
In this case, the cross-sectional area at a given value of x is given by the difference in the areas between the curves f(x) and g(x) at that value of x. So the cross-sectional area is:
A(x) = π[f(x)^2 - g(x)^2]
The limits of integration are from 0 to a, as given in the question.
Therefore, the volume becomes:
V = ∫[0,a] π[f(x)^2 - g(x)^2] dx
To simplify the calculation, we need to express f(x) and g(x) in terms of x, which gives:
V = ∫[0,a] π[(x^2)^2 - (2x)^2] dx
= ∫[0,a] π[x^4 - 4x^2] dx
= π ∫[0,a] (x^4 - 4x^2) dx
Integrating this expression, we have:
V = π [(x^5)/5 - (4x^3)/3] evaluated from 0 to a
= π [(a^5)/5 - (4a^3)/3] - π [(0^5)/5 - (4(0^3))/3]
= π [(a^5)/5 - (4a^3)/3]
Now, we need to set this equal to 7πa^3/12, as stated in the question:
π [(a^5)/5 - (4a^3)/3] = 7πa^3/12
To solve for a, we can cancel out the π terms and simplify the equation:
(a^5)/5 - (4a^3)/3 = 7a^3/12
Multiplying through by the least common denominator (LCD) of 60 to eliminate fractions, we have:
12(a^5) - 20(4a^3) = 5(7a^3)
12a^5 - 80a^3 = 35a^3
12a^5 - 80a^3 - 35a^3 = 0
12a^5 - 115a^3 = 0
Factoring out an a^3 term:
a^3(12a^2 - 115) = 0
Now, set each factor equal to zero and solve for a:
a^3 = 0 --> a = 0
12a^2 - 115 = 0
Solving for a^2:
12a^2 = 115
a^2 = 115/12
Taking the square root:
a = ±√(115/12)
Since we're looking for a value between 0 and 2, we discard the negative solution and take the positive square root:
a = √(115/12)
So, the value of a between 0 and 2, such that the volume of the solid produced when the region between the graphs of f and g is revolved about the y-axis is 7πa^3/12, is a = √(115/12).