Washer Method

Silvanoshei

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Feb 18, 2013
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Finding the volumes by revolving the regions bounded by the lines and curves about the x-axis.

y=x,y=1,x=0\displaystyle y=x, y=1, x=0

So...

π01[x]2dx\displaystyle \pi ∫_{0}^{1}\left[x\right]^2dx

π01x2dx\displaystyle \pi ∫_{0}^{1}x^2dx

π01x33\displaystyle \pi ∫_{0}^{1}\frac{x^{3}}{3}

π[x33]@1=π3\displaystyle \pi\left[\frac{x^{3}}{3}\right] @ 1 = \frac{\pi}{3}

π[x33]@0=0\displaystyle \pi\left[\frac{x^{3}}{3}\right] @ 0 = 0

π30=π3\displaystyle \frac{\pi}{3} - 0 = \frac{\pi}{3}

Book gives an answer of: 2π3?\displaystyle \frac{2\pi}{3}?
 
Finding the volumes by revolving the regions bounded by the lines and curves about the x-axis.

y=x,y=1,x=0\displaystyle y=x, y=1, x=0

So...

π01[x]2dx\displaystyle \pi ∫_{0}^{1}\left[x\right]^2dx

π01x2dx\displaystyle \pi ∫_{0}^{1}x^2dx

π01x33\displaystyle \pi ∫_{0}^{1}\frac{x^{3}}{3}

π[x33]@1=π3\displaystyle \pi\left[\frac{x^{3}}{3}\right] @ 1 = \frac{\pi}{3}

π[x33]@0=0\displaystyle \pi\left[\frac{x^{3}}{3}\right] @ 0 = 0

π30=π3\displaystyle \frac{\pi}{3} - 0 = \frac{\pi}{3}

Book gives an answer of: 2π3?\displaystyle \frac{2\pi}{3}?

Did you sketch the area of revolution?

Volume of washer should be = π * (12-y2) dx
 
Finding the volumes by revolving the regions bounded by the lines and curves about the x-axis.

y=x,y=1,x=0\displaystyle y=x, y=1, x=0

So...

π01[x]2dx\displaystyle \pi ∫_{0}^{1}\left[x\right]^2dx

π01x2dx\displaystyle \pi ∫_{0}^{1}x^2dx

π01x33\displaystyle \pi ∫_{0}^{1}\frac{x^{3}}{3}

π[x33]@1=π3\displaystyle \pi\left[\frac{x^{3}}{3}\right] @ 1 = \frac{\pi}{3}

π[x33]@0=0\displaystyle \pi\left[\frac{x^{3}}{3}\right] @ 0 = 0

π30=π3\displaystyle \frac{\pi}{3} - 0 = \frac{\pi}{3}

Book gives an answer of: 2π3?\displaystyle \frac{2\pi}{3}?

You are finding the volume of the wrong region. It is the little triangle formed above y=x, below y = 1 and to the right of the y-axis (x=0). Thus you have the washer method.

π01(12x2)dx\displaystyle \pi ∫_{0}^{1}(1^2-x^2)dx

Can you take it from here?
 
Oh my, such a small error.

So...


π(1133)(0)=π23\displaystyle \pi(1-\frac{1^{3}}{3})-(0) = \frac{\pi2}{3}

Blah, thanks guys! :cool:
 
Be careful: what you wrote, π23\displaystyle \dfrac{\pi2}{3}, could easily be mistaken for π23\displaystyle \dfrac{\pi^2}{3}. Better is 2π3\displaystyle \dfrac{2\pi}{3}.
 
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