Find the volume of the solid of revolution formed by rotating the region bounded by $y=x^3$ and $y=x$ about the line $x=-2$. Use the shell method. V = [ ? ] Round your answer to the nearest thousandth.
Answer (2)
Find the intersection points of y = x 3 and y = x , which are x = − 1 , 0 , 1 .
Set up the volume integral using the shell method: V = 2 π ∫ − 1 1 ( x + 2 ) ∣ x − x 3 ∣ d x .
Split the integral into two parts due to the absolute value and symmetry: V = 2 π [ ∫ − 1 0 ( x + 2 ) ( x 3 − x ) d x + ∫ 0 1 ( x + 2 ) ( x − x 3 ) d x ] .
Evaluate the integrals and find the volume: V = 2 π ≈ 6.283 .
Explanation
Problem Setup We are asked to find the volume of the solid of revolution formed by rotating the region bounded by y = x 3 and y = x about the line x = − 2 using the shell method. We need to find the intersection points of the curves and set up the integral for the volume.
Finding Intersection Points First, let's find the intersection points of the curves y = x 3 and y = x . We set x 3 = x , which gives x 3 − x = 0 . Factoring, we get x ( x 2 − 1 ) = 0 , so x ( x − 1 ) ( x + 1 ) = 0 . Thus, the intersection points are x = − 1 , 0 , 1 .
Setting up the Integral The shell method formula for the volume of revolution about a vertical axis is given by V = 2 π ∫ a b r ( x ) h ( x ) d x , where r ( x ) is the radius of the shell and h ( x ) is the height of the shell. In this case, the axis of rotation is x = − 2 , so the radius of the shell is r ( x ) = x − ( − 2 ) = x + 2 . The height of the shell is the difference between the two curves, which is h ( x ) = ∣ x − x 3 ∣ .
Using Symmetry Since the region is symmetric with respect to the y-axis, we can write the volume as
V = 2 π ∫ − 1 1 ( x + 2 ) ∣ x − x 3 ∣ d x = 2 π [ ∫ − 1 0 ( x + 2 ) ( x 3 − x ) d x + ∫ 0 1 ( x + 2 ) ( x − x 3 ) d x ]
Evaluating the Integrals Now, we evaluate the integrals:
∫ − 1 0 ( x + 2 ) ( x 3 − x ) d x = ∫ − 1 0 ( x 4 − x 2 + 2 x 3 − 2 x ) d x = [ 5 x 5 − 3 x 3 + 4 2 x 4 − x 2 ] − 1 0 = 0 − ( 5 − 1 − 3 − 1 + 2 1 − 1 ) = − ( − 5 1 + 3 1 + 2 1 − 1 ) = − ( 30 − 6 + 10 + 15 − 30 ) = − ( 30 − 11 ) = 30 11
∫ 0 1 ( x + 2 ) ( x − x 3 ) d x = ∫ 0 1 ( x 2 − x 4 + 2 x − 2 x 3 ) d x = [ 3 x 3 − 5 x 5 + x 2 − 4 2 x 4 ] 0 1 = 3 1 − 5 1 + 1 − 2 1 = 30 10 − 6 + 30 − 15 = 30 19
Calculating the Volume So, the volume is
V = 2 π ( 30 11 + 30 19 ) = 2 π ( 30 30 ) = 2 π
Final Answer The volume of the solid of revolution is 2 π ≈ 6.283 .
Examples
Understanding volumes of revolution is crucial in many engineering applications. For example, when designing tanks or containers, calculating the volume accurately ensures that the container can hold the required amount of liquid or gas. The shell method, as used here, is particularly useful when the axis of rotation is parallel to the axis of integration, making it easier to set up and solve the integral. This ensures efficient and cost-effective designs in various industries.
To find the volume of the solid formed by rotating the region bounded by y = x 3 and y = x around the line x = − 2 using the shell method, we find the intersection points at x = − 1 , 0 , 1 . The calculated volume is V = 2 π × 1 = 2 π ≈ 6.283 after evaluating the necessary integrals.
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