Question

Where am I going wrong?(Calc). THe question is: Find the volume of the triangle enclosed by the vertices (3,2)(3,4)(7,4) about the y-axis. What I have so far is the r(y)= the constant so 3. Then it tells me to make the equation of the line between the point (3,2)(7,4) with the formula y-y1/x-x1= y2-y1/x2-x1..which gave me 1/x-3=1. I plugged that into the integral and the answer is wrong. Can anyone help?

Answer 1

what exactly does your integral look like that you tried?

Example:
Int from a to b of (whatever equation) dx

Answer 2

top limit i have as 4, lower as 3

Answer 3

then i was told from the lab to use the equation i get from that formula, but i have yet to get that right, then after that the constant squared, and the whole integral * pi

Answer 4

i.e example (6y-4)^2-3^2 dy

Answer 5

i can't seem to get the 6y-4(example) part correct

Answer 6

ok, thank you for the details, that will help us give you a solution

Answer 7

and just to be clear you are getting
\[\frac{1}{x-3} = 1\] as your equation?

Answer 8

yes. i have y as 3, so y-y1/x-x1 gave me 3-2/x-3, and y2-y1/x2-x1 gave me 4-2/4-2, so 1/x-3=1 is what im stuck with

Answer 9

aha! there's the issue, you shouldn't be plugging in anything for y in the first part; you should not get 3 - 2

Answer 10

that is an arbitrary y just like the x below it

Answer 11

so 0-6/3-3= 0... and 1? 0 does not equal 1? lost

Answer 12

If we collect the data this way it becomes more clear:
(x1,y1) = (3,2) (x2,y2) = (7,4)

The formula is (y - y1)/(x - x1) = (y2 - y1)/(x2 - x1) where y and x are unspecified to give you an actual equation

Answer 13

You only need to plug in for x1, y1, x2, and y2.

Answer 14

This should give you \[\frac{y - 2}{x - 3} = \frac{4 - 2}{7 - 3}\]

Answer 15

Do you see that? :)

Answer 16

yeah

Answer 17

OK, so what would you think to do from here?

Answer 18

y-2/x-3=1/2 then

Answer 19

good, and then? :)

Answer 20

if we want the line equation we need y=x so multiply the x-3 on both sides?

Answer 21

ok, sounds good

Answer 22

y= (1/2)(x-1)

Answer 23

1/2x -1 is our equation?

Answer 24

actually, hold up, is that supposed to be a minus inside the parentheses?

Answer 25

where?

Answer 26

We had y - 2 = (1/2)(x - 3), which becomes y - 2 = (1/2)x - 3/2, which becomes y = (1/2)x + 1/2 after adding 2 to both sides.

Answer 27

it should be a plus sign is all I'm saying.

Answer 28

so that gives my integral as: (pi) [integral] UPPER: 4 LOWER:3 -> (1/2y+1/2)^2-(3)^2 dy ?

Answer 29

OK, got to be careful here, you can't just change the x's into y's like that. However, you are correct in trying to get only y's inside. Can you solve the equation of the line in terms of y's instead?

Answer 30

How about taking y = (1/2)x + 1/2 and getting, y - 1/2 = (1/2)x, then, 2y - 1 = x. Does that make sense to you?

Answer 31

oh crap. yeah

Answer 32

so 82pi/3?

Answer 33

lol, this is a tricky question... and the negative area should have scared you anyway. :)

Answer 34

checking...

Answer 35

hmm i did something wrong

Answer 36

are my limits correct? 4/3?

Answer 37

the integral should be \[\pi \int\limits_3^4 (2y-1)^2 - 3^2 dy\]

Answer 38

thats what i have. isnt that 82/3? then with the pi 82pi/3?

Answer 39

I'm almost there...

Answer 40

yeah, that looks correct to me... is the answer in the book different?

Answer 41

let me make sure i got all the numbers correct

Answer 42

oh wait, the limits are 2 to 4 not 3!

Answer 43

my bad, sorry on that

Answer 44

because (3,2) has a y of 2 and (7,4) has a y of 4

Answer 45

ha yeah thats it :)

Answer 46

cool, awesome!

Answer 47

got it right, thanks. 7th time is the wonder

Answer 48

much appreciated.

Answer 49

lol, no but seriously it takes like 10 times on these calc questions sometimes. :) Good job!

Answer 50

cya in the future, thanks again

Answer 51

take care