Can you please help me understand how the answer of this question is (7π /15)?
The question is: Let R be the region bounded by y=x^2, y=0, and x=1. Compute the volume of the solid formed by revolving R about the line y=1.
I know that V=π (integral sign) A(x)^2 dx.
So I want to set it up as V= π (int) sqrt(x) dx with interval [1,0]. However, it's not working out. Can somebody please enlighten me?
2007-12-01 13:34:37 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Use the washer method to integrate.
Once you draw the graph you can see that you need to integrate over the inverval [0, 1].
Volume = ∫[πR² - πr²] dx = π∫[R² - r²] dx
= π∫[(1 - 0)² - (1 - x²)²] dx
= π∫[1 - (1 - 2x² + x^4)] dx
= π∫(2x² - x^4) dx
= π(2x³/3 - x^5/5) | [Evaluated from 0 to 1]
= π(2/3 - 1/5) - 0 = 7π/15
2007-12-01 13:56:39 · answer #1 · answered by Northstar 7 · 1⤊ 0⤋
The easiest way to do this is to make this a function of y, get rid of the exponent by taking the natural logarithm of both sides, and then implicitly differentiate. y = x^x ln y = x ln x (The exponent now goes in front of the logarithm function.) Now, implicitly differentiate. 1/y dy/dx = ln x + x(1/x) Simplify it a bit and multiply both sides by y. dy/dx = (ln x + 1)y You could leave it like that, but you probably want to get the answer in terms of only x, so recall what y originally is (x^x), and substitute it in there. dy/dx = (ln x + 1)x^x = (ln x)x^x + x^x
2016-05-27 04:28:15 · answer #2 · answered by ? 3 · 0⤊ 0⤋