R Run hard@thehsc Well-Known Member Joined Oct 7, 2021 Messages 784 Gender Male HSC 2022 Apr 23, 2022 #1 For the following volume of revolution question (part d), what did you guys get: I think I got 16pi/105 unit cubed. Can someone confirm this? Thanks
For the following volume of revolution question (part d), what did you guys get: I think I got 16pi/105 unit cubed. Can someone confirm this? Thanks
C cossine Well-Known Member Joined Jul 24, 2020 Messages 627 Gender Male HSC 2017 Apr 24, 2022 #2 Run hard@thehsc said: For the following volume of revolution question (part d), what did you guys get: View attachment 35498 I think I got 16pi/105 unit cubed. Can someone confirm this? Thanks Click to expand... x(x^2-1) = x(x+1)(x-1) bounds from -1 to 1 pi integral (x^3 -x)^2 dx What your taking is the sum of an infinite number circles across the x-axis Integral Calculator • With Steps! Solve definite and indefinite integrals (antiderivatives) using this free online calculator. Step-by-step solution and graphs included! www.integral-calculator.com Can you confirm the answer
Run hard@thehsc said: For the following volume of revolution question (part d), what did you guys get: View attachment 35498 I think I got 16pi/105 unit cubed. Can someone confirm this? Thanks Click to expand... x(x^2-1) = x(x+1)(x-1) bounds from -1 to 1 pi integral (x^3 -x)^2 dx What your taking is the sum of an infinite number circles across the x-axis Integral Calculator • With Steps! Solve definite and indefinite integrals (antiderivatives) using this free online calculator. Step-by-step solution and graphs included! www.integral-calculator.com Can you confirm the answer
mmmmmmmmaaaaaaa Well-Known Member Joined Jun 11, 2021 Messages 1,416 Gender Male HSC 2022 Apr 24, 2022 #3 16/105 pi is correct