Washer Method

[Silvanoshei]

Finding the volumes by revolving the regions bounded by the lines and curves about the x-axis.

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

So...

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

\(\displaystyle \pi ∫_{0}^{1}x^2dx\)

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

\(\displaystyle \pi\left[\frac{x^{3}}{3}\right] @ 1 = \frac{\pi}{3}\)

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

\(\displaystyle \frac{\pi}{3} - 0 = \frac{\pi}{3}\)

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

 

[Deleted member 4993]

Finding the volumes by revolving the regions bounded by the lines and curves about the x-axis.

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

So...

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

\(\displaystyle \pi ∫_{0}^{1}x^2dx\)

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

\(\displaystyle \pi\left[\frac{x^{3}}{3}\right] @ 1 = \frac{\pi}{3}\)

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

\(\displaystyle \frac{\pi}{3} - 0 = \frac{\pi}{3}\)

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

Did you sketch the area of revolution?

Volume of washer should be = π * (1²-y²) dx

 

[srmichael]

Finding the volumes by revolving the regions bounded by the lines and curves about the x-axis.

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

So...

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

\(\displaystyle \pi ∫_{0}^{1}x^2dx\)

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

\(\displaystyle \pi\left[\frac{x^{3}}{3}\right] @ 1 = \frac{\pi}{3}\)

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

\(\displaystyle \frac{\pi}{3} - 0 = \frac{\pi}{3}\)

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

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.

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

Can you take it from here?

 

[Silvanoshei]

Oh my, such a small error.

So...

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

Blah, thanks guys! :cool:

 

[HallsofIvy]

Be careful: what you wrote, \(\displaystyle \dfrac{\pi2}{3}\), could easily be mistaken for \(\displaystyle \dfrac{\pi^2}{3}\). Better is \(\displaystyle \dfrac{2\pi}{3}\).