Washer Method

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kneaiak

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Jan 11, 2013
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"Use the washer method to find the volume of the solid generated when the region bounded by y=16x and y=8x^2 - x^3 is revolved about the x-axis."


int 0 to 4 (pi) ((16x)^2 - (8x^2 - x^3)^2) dx

(pi) int 0 to 4 of 256x^2 -64x^4 + 16x^5 - x^6 dx

(pi) int 0 to 4 of -x^6 + 16x^5 - 64x^4 + 256x^2 dx

integration:

(pi) (-((x^7)/7) + ((16x^6)/6) - ((64x^5)/5) + ((256x^3)/3)) evaluated from 0 to 4

When I plug in my upper bound of 4 I get (2944/35)(pi) and my lower bound of 0 cancels everything out.

Finally..... (2944/35)(pi)

This seems very odd

Thanks for any help.
 
kneaiak said:
"Use the washer method to find the volume of the solid generated when the region bounded by y=16x and y=8x^2 - x^3 is revolved about the x-axis."


int 0 to 4 (pi) ((16x)^2 - (8x^2 - x^3)^2) dx

(pi) int 0 to 4 of 256x^2 -64x^4 + 16x^5 - x^6 dx

(pi) int 0 to 4 of -x^6 + 16x^5 - 64x^4 + 256x^2 dx

integration:

(pi) (-((x^7)/7) + ((16x^6)/6) - ((64x^5)/5) + ((256x^3)/3)) evaluated from 0 to 4

When I plug in my upper bound of 4 I get (2944/35)(pi) and my lower bound of 0 cancels everything out.

Finally..... (2944/35)(pi)

This seems very odd

Thanks for any help.
Click to expand...
Rotation about the x-axis by the washer method:
\(\displaystyle \displaystyle \int_0^4 \pi \left((16x)^2 - (8x^2 - x^3)^2\right)dx\)
\(\displaystyle \displaystyle = \pi \left[-\frac{x^7}{7} + \frac{4^2 x^6}{6} - \frac{4^3 x^5}{5} + \frac{4^4 x^3}{3}\right]_0^4\)

This is the same as what you had, with the numerators all expressed as powers of 4. When x=4, that lets me take a factor of 4^8 outside the brackets:

\(\displaystyle \displaystyle = \pi 4^8 \left[-\frac{1}{28} + \frac{1}{6} - \frac{1}{5} + \frac{1}{12} \right]\)

\(\displaystyle \displaystyle = \pi 4^8 / 70 = (32768/35) \pi \)

Not quite the same number you got. We had better both check!
 
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