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 (1)
Find the intersection points of y = x 3 and y = x , which are x = − 1 , 0 , 1 .
Set up the integral for the volume using the shell method: V = 2 π ∫ − 1 1 ( x + 2 ) ∣ x − x 3 ∣ d x .
Evaluate the integral by splitting it into two parts: 2 π ( ∫ − 1 0 ( x + 2 ) ( x 3 − x ) d x + ∫ 0 1 ( x + 2 ) ( x − x 3 ) d x ) .
Calculate 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.
Finding Intersection Points First, we need to find the intersection points of the curves y = x 3 and y = x . We set x 3 = x , which gives us x 3 − x = 0 . Factoring, we have x ( x 2 − 1 ) = x ( x − 1 ) ( x + 1 ) = 0 . Thus, the intersection points occur at x = − 1 , 0 , 1 .
Setting up the Shell Method Since we are rotating about the vertical line x = − 2 , the radius of a cylindrical shell is given by r = x − ( − 2 ) = x + 2 . The height of the cylindrical shell is the difference between the y values of the two curves. For − 1 ≤ x ≤ 0 , we have x 3 ≥ x , so the height is x 3 − x . For 0 ≤ x ≤ 1 , we have x ≥ x 3 , so the height is x − x 3 .
Setting up the Integral The volume of a cylindrical shell is given by d V = 2 π r h d x = 2 π ( x + 2 ) h d x . The total volume is the sum of the volumes of the cylindrical shells, which can be expressed as the integral:
V = ∫ − 1 1 2 π ( 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 + 2 x 3 − x 2 − 2 x ) d x = [ 5 x 5 + 4 2 x 4 − 3 x 3 − x 2 ] − 1 0 = 0 − ( − 5 1 + 2 1 + 3 1 − 1 ) = 5 1 − 2 1 − 3 1 + 1 = 30 6 − 15 − 10 + 30 = 30 11
∫ 0 1 ( x + 2 ) ( x − x 3 ) d x = ∫ 0 1 ( x 2 + 2 x − x 4 − 2 x 3 ) d x = [ 3 x 3 + x 2 − 5 x 5 − 4 2 x 4 ] 0 1 = 3 1 + 1 − 5 1 − 2 1 = 30 10 + 30 − 6 − 15 = 30 19
Calculating the Volume Therefore, the volume is:
V = 2 π ( 30 11 + 30 19 ) = 2 π ( 30 30 ) = 2 π
Final Answer Finally, we approximate the volume to the nearest thousandth:
V = 2 π ≈ 6.283
Examples
Understanding volumes of revolution is crucial in many engineering applications. For instance, when designing tanks or containers, calculating the volume helps determine the capacity and material requirements. In medical device manufacturing, knowing the precise volume of components is essential for drug delivery systems. This problem showcases how calculus can be applied to real-world design and manufacturing processes.
