Question

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width.
y=7x^2
y^2 = (1/7)x
find area of the region

Answer 1

this sounds familiar ;)

Answer 2

helloo and yes D: i feel like i did all the steps right, but its not the right answer >< i set both the equations to eachother to find the boundaries, thats the first step right?

Answer 3

7x^2 = sqrt of (1/7)x
i square both sides to get rid of the radical so im left with
7x^4 = (1/7)x
i multiply both sides by 7, i get
49x^4 = x
subtract x from both sides, you get
49x^4 - x = 0

Answer 4

can we double int it?

Answer 5

i believe so

Answer 6

y = sqrt(1/7x) y=7x^2

{SS} dy dx ; x=[0,1]; y = [7x^2,sqrt(1/7x)] right?

Answer 7

{S} sqrt(1/7x) - 7x^2 dx ; [0,1]

Answer 8

how did you get 0 and 1 ?

Answer 9

sqrt(1/7) 2x sqrt(x/3) - 7/x^3/3

Answer 10

thats where your graphs intersect at

Answer 11

to find the intersection, you set the equations equal to eachother right?

Answer 12

2sqrt(1/21)-7/3 i believe is the area

Answer 13

yes

Answer 14

did i miss that up lol

Answer 15

thats what i did at first, but i got x = 0 and x = cube root of 1/49

Answer 16

y= 7x^2; and x=7y^2

Answer 17

i could be wrong, trying tis in my head lol

Answer 18

lol youre right, but wouldnt we have to set that second equation that y is on one side by itself, not x ?

Answer 19

y=sqrt(1/7x) ; y = 7x^2

1/7x = 49x^4

49 x^4 - 1/7x = 0

x(49x^3 - 1/7) = 0 x=0 to.....my brain shut off lol

Answer 20

1/7x(343x^3 - 1) = 0

Answer 21

x = 0 and x = cube root of 1/343 ?

Answer 22

wolfram tells me that (1,1/7) is the solution to these; lets try that out :)

Answer 23

sqrt(1/7*1) = sqrt(1/7)

7*1 = 7....... ack!!

Answer 24

lol xD

Answer 25

hmm

Answer 26

deep breath....... ok

Answer 27

y=7x^2 and y^2 = 1/7x right?

Answer 28

yup

Answer 29

we know zero is a good one

Answer 30

yes

Answer 31

y =1 1 7
---- = 1/7; at x=1/7, y = 1/7 ; thats y=x^2
7 7

Answer 32

at x = 1/7; we get y = sqrt(1/7 1/7) = sqrt(1/49) = 1/7

Answer 33

so lets int this thing from 0 to 1/7 :)

Answer 34

im sorry im a little lost, how did you get 1/7 for x ?

Answer 35

[S] 7x^2 - sqrt(1/7x) dx ; [0,1/7] ....i guessed and was right :)

Answer 36

xD awesome haha

Answer 37

i had an inclination, but couldnt verify it till i just plugged it in ;)

Answer 38

i know that this is 2 parabolas at right angles to each other; so they only meet at 2 point, to that one works and has to be it lol

Answer 39

7x^3/3 - 2sqrt((7x)^3)/3 ??

Answer 40

is the 2aqrt((7x)^3)/3 the same as 2x^(3/2) divided by 21 ? cuz i put sqrt of 1/7 as (1/7)^(1/2) and found the antiderivative like that

Answer 41

sqrt(1/7)* sqrt(x) = sqrt(1/7)* 2sqrt(x^3)/3

7x^3/3 - 2sqrt(1/7x^3)/3 right?

Answer 42

the sqrt(1/7) is actually a constant, so we can pull it out of the way and just do sqrt(x)

Answer 43

sqrt(ab) = sqrt(a) * sqrt(b)

Answer 44

ohhh i see

Answer 45

1/49(3) - 2sqrt(1/7^4)/3 should be the area

Answer 46

how comes its (1/7^4) ?

Answer 47

1 2sqrt(1/2401)
--- - ------------
147 3

(1/7) * (1/7)^3 = 1/7^4

Answer 48

as long as i integrated it right, and im pretty sure i did :)

Answer 49

dbl check it tho ;)

Answer 50

im going over it right now :D

Answer 51

\[\int\limits_{} (7x^2 - \sqrt{\frac{1}{7}x} )dx\]

Answer 52

http://www.wolframalpha.com/input/?i=int%287x^2+-sqrt%28x%2F7%29%29dx++from+0+to+1%2F7

Answer 53

id take the absolute value of that :)

Answer 54

thats what i got too :) and i jus ttake the absolute value of that since its negative right ?

Answer 55

correct; just means we got the wrong order

Answer 56

see... http://www.wolframalpha.com/input/?i=int%28sqrt%28x%2F7%29-7x^2%29dx++from+0+to+1%2F7

Answer 57

gotcha. thank you once again. always coming to my rescue :D haha youre the best !