Question

Using the disk method find the volume of the area bounded by y = the square root of (x - 1) y = (x-1)/2 rotated around the x axis. I'll show you what I wrote in the next answer.

Answer 1

So what I did was first find the limits of integration by setting the two y values equal to each other. Like so.
.5(x-1) - (x-1)^1/2 = 0

I got x = 5 and x =1 for my x values.
So I set my upper limit to 5, and my lower limit to 1.

Answer 2

I then wrote \[\int\limits_{1}^{5} \frac{ (x-1) }{ 2 }^{2} - (x-1)\]
and took the anti derivative of that which got me
\[(\frac{ x-1 }{ 6 })^{3} - \frac{ (x-1) }{ 2 }^{2}\]

When I solved for that I got (.29-4) -(0)
which gave me a negative volume.

Which step did I do incorrectly?

Answer 3

the integrals. take them separately and you'll see

\[\int\limits_{1}^{5}[(x-1)^{1/2} - x + 1]dx\]

Answer 4

Huh? Where did the .5? Go.
The original equation was.

\[\frac{ (x-1) }{ 2 }\] and \[\sqrt{x-1}\]

I just rewrote them. Are you saying I should redo the integral step? Let's see what I get.

Answer 5

my bad. add it

Answer 6

I can just square both, it's part of the formula when using the disk method to find volume.

Answer 7

\[∫\frac{ (x−1) }{ 2 } dx=\frac{ 1 }{ 2 } \left(\frac{ x^2 }{ 2 }−x \right)≠\frac{ 1 }{ 2 }(x^2−2x+1)\]

Answer 8

sorry for earlier. forgot how to do the actual disk integration so i focused on that

Answer 9

instead of this

Answer 10

make that dne 1/4(x-1)^2

Answer 11

I still have no clue what you are talking about or no clue where I made my mistake. It seems like everything you are saying has nothing to do at all with what I am asking.

Answer 12

yes

Answer 13

second integral

Answer 14

\[\int\limits_{1}^{5}[\frac{ (x-1)^2 }{ 4 }−(x−1)]dx\]

Answer 15

except not 1/2 lol my bad

Answer 16

\[\int\limits_{}^{}(x-1)dx = \frac{ x^2 }{ 2 } - x \neq \frac{ 1 }{ 2 }(x^2 -2x + 1)\]

Answer 17

So what you are trying to say is that the anti derivative of (x-1) is NOT (x-1)^2/2?

Answer 18

yes

Answer 19

Well what is it then? Do you mean that I should not get rid of the square root untill after I have taken the anti derivative? If I don't get rid of the square root I get a non real answer.

Answer 20

So iSo instead it would be

Answer 21

no just

\[\int\limits_{}^{}(x-1)dx = \frac{ x^2 }{ 2 } - x \neq \frac{ (x-1)^2 }{ 2 }\]

integral is fine