Volume - Washer Method

[kneaiak]

"Use the Washer Method to find the volume of the solid generated when the region bounded by y=16x and y=8xsquared-xcubed is revolved about the x-axis."

I found my integration points to be 0 (lower) to 4 (upper).

int from 0 to 4 of (pi)(R squared - r squared) dx

int from 0 to 4 of (pi)((16xsquared - (8xsquared - xcubed)) dx

(pi) int from 0 to 4 of 256xsquared - (64xforth- 2xcubed + xsixth) dx

(pi) int from 0 to 4 of 256xsquared - 64xforth + 2xcubed - sixth) dx

I finally integrate and come up with the below which makes no sense because I would end up with negative volume......

(pi) ((256xcubed)/3)-((64xfifth)/5)+((2xforth)/4)-(xseventh/7) evaluated from o to 4

once I apply the numbers to get a common denominator and plug in my upper limit of 4 I get a negative number...Impossible.

Any ideas where I went wrong?

 

[HallsofIvy]

"Use the Washer Method to find the volume of the solid generated when the region bounded by y=16x and y=8xsquared-xcubed is revolved about the x-axis.

Better to use "^" to indicate powers: y= 8x^2- x^3 is easier to read.

I found my integration points to be 0 (lower) to 4 (upper).

int from 0 to 4 of (pi)(R squared - r squared) dx

int from 0 to 4 of (pi)((16xsquared - (8xsquared - xcubed)) dx

Use parentheses: (16x)^2 (or (16x)squared) and you don't have the "square" on (8x^2- x^3)

(pi) int from 0 to 4 of 256xsquared - (64xforth- 2xcubed + xsixth) dx

(8x^2- x^3)^2= 64x^4- 16x^5+ x^6. You seem to have forgotten to multiply by "8x^2" in the middle term.
I think this is the error that gives you the wrong answer.

(pi) int from 0 to 4 of 256xsquared - 64xforth + 2xcubed - sixth) dx

I finally integrate and come up with the below which makes no sense because I would end up with negative volume......

(pi) ((256xcubed)/3)-((64xfifth)/5)+((2xforth)/4)-(xseventh/7) evaluated from o to 4

once I apply the numbers to get a common denominator and plug in my upper limit of 4 I get a negative number...Impossible.

Any ideas where I went wrong?

 

[MarkFL]

You have incorrectly expanded the integrand. It appears you have the correct integral:

\(\displaystyle \displaystyle V=\pi\int_0^4(16x)^2-(8x^2-x^3)^2\,dx\)

Now, we may rewrite the integrand as follows:

\(\displaystyle (16x)^2-(8x^2-x^3)^2=256x^2-x^4(8-x)^2=256x^2-x^4(64-16x+x^2)=-x^6+16x^5-64x^4+256x^2\)

You simply squared the binomial incorrectly, making a minor error. Your line:

(pi) int from 0 to 4 of 256xsquared - (64xforth- 2xcubed + xsixth) dx

should be:

(pi) int from 0 to 4 of 256xsquared - (64xforth- 16xcubed + xsixth) dx

Do you see why we need 16 instead of 2 there?

 

[kneaiak]

yep, now i see it, looks like I'm trying to go too quickly and not double checking my work. Thanks again.

 

[MarkFL]

Also, HallsofIvy raises a very good point (as he usually does :cool. Where you have written:

int from 0 to 4 of (pi)((16xsquared - (8xsquared - xcubed)) dx

you should use parentheses around terms raised to a power (such as 16x) and show that the cubic curve is also being squared:

int from 0 to 4 of (pi)((16x)squared - (8xsquared - xcubed)squared) dx

I assumed from your previous and subsequent lines that this was simply an unintentional omission.

For even more clarity, for exponents use the caret character "^" as follows:

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

 

[kneaiak]

Still screwing something up, I come up with ((3735552(pi))/315) after plugging in my upper limit of 4.....am I forgetting a way to reduce to powers? My numbers seem to be blown up

 

[kneaiak]

Thanks for "^" tip too everyone. Thanks so much for all the help, I'd be shot without the forum.

 

[MarkFL]

We have:

\(\displaystyle \displaystyle V=\pi\int_0^4(-x^6+16x^5-64x^4+256x^2)\,dx\)

\(\displaystyle \displaystyle V=\pi\left[-\frac{x^7}{7}+\frac{8x^6}{3}-\frac{64x^5}{5}+\frac{256x^3}{3} \right]_0^4\)

\(\displaystyle \displaystyle V=\pi\left(-\frac{4^7}{7}+\frac{8\cdot4^6}{3}-\frac{64\cdot4^5}{5}+\frac{256\cdot4^3}{3} \right)\)

\(\displaystyle \displaystyle V=\pi\left(-\frac{2^{14}}{7}+\frac{2^{15}}{3}-\frac{2^{16}}{5}+\frac{2^{14}}{3} \right)\)

\(\displaystyle \displaystyle V=2^{14}\pi\left(-\frac{1}{7}+\frac{2}{3}-\frac{4}{5}+\frac{1}{3} \right)\)

\(\displaystyle \displaystyle V=2^{14}\pi\left(\frac{2}{35} \right)=\frac{2^{15}\pi}{35}=\frac{32768\pi}{35}\)