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anonymous
 5 years ago
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
anonymous
 5 years ago
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

This Question is Closed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0this sounds familiar ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0helloo 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.07x^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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0can we double int it?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y = sqrt(1/7x) y=7x^2 {SS} dy dx ; x=[0,1]; y = [7x^2,sqrt(1/7x)] right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0{S} sqrt(1/7x)  7x^2 dx ; [0,1]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get 0 and 1 ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt(1/7) 2x sqrt(x/3)  7/x^3/3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0thats where your graphs intersect at

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0to find the intersection, you set the equations equal to eachother right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02sqrt(1/21)7/3 i believe is the area

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0did i miss that up lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats what i did at first, but i got x = 0 and x = cube root of 1/49

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i could be wrong, trying tis in my head lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol youre right, but wouldnt we have to set that second equation that y is on one side by itself, not x ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y=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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x = 0 and x = cube root of 1/343 ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0wolfram tells me that (1,1/7) is the solution to these; lets try that out :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt(1/7*1) = sqrt(1/7) 7*1 = 7....... ack!!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0deep breath....... ok

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y=7x^2 and y^2 = 1/7x right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we know zero is a good one

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y =1 1 7  = 1/7; at x=1/7, y = 1/7 ; thats y=x^2 7 7

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0at x = 1/7; we get y = sqrt(1/7 1/7) = sqrt(1/49) = 1/7

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so lets int this thing from 0 to 1/7 :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im sorry im a little lost, how did you get 1/7 for x ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0[S] 7x^2  sqrt(1/7x) dx ; [0,1/7] ....i guessed and was right :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i had an inclination, but couldnt verify it till i just plugged it in ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.07x^3/3  2sqrt((7x)^3)/3 ??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt(1/7)* sqrt(x) = sqrt(1/7)* 2sqrt(x^3)/3 7x^3/3  2sqrt(1/7x^3)/3 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the sqrt(1/7) is actually a constant, so we can pull it out of the way and just do sqrt(x)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt(ab) = sqrt(a) * sqrt(b)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01/49(3)  2sqrt(1/7^4)/3 should be the area

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how comes its (1/7^4) ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01 2sqrt(1/2401)    147 3 (1/7) * (1/7)^3 = 1/7^4

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0as long as i integrated it right, and im pretty sure i did :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im going over it right now :D

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{} (7x^2  \sqrt{\frac{1}{7}x} )dx\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=int%287x^2+sqrt%28x%2F7%29%29dx++from+0+to+1%2F7

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0id take the absolute value of that :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats what i got too :) and i jus ttake the absolute value of that since its negative right ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0correct; just means we got the wrong order

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0see... http://www.wolframalpha.com/input/?i=int%28sqrt%28x%2F7%297x^2%29dx++from+0+to+1%2F7

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0gotcha. thank you once again. always coming to my rescue :D haha youre the best !
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