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anonymous
 one year ago
hello, advance definite integral area bound by 4 curves.
https://fbcdnsphotosea.akamaihd.net/hphotosakxpl1/v/t34.012/11939020_821736254611990_2090835384_n.jpg?oh=3cd85831dc162d90184347e50a890cc8&oe=55DD3400&__gda__=1440572593_c4c4257c8f45c3517dc933a3312f81db
anonymous
 one year ago
hello, advance definite integral area bound by 4 curves. https://fbcdnsphotosea.akamaihd.net/hphotosakxpl1/v/t34.012/11939020_821736254611990_2090835384_n.jpg?oh=3cd85831dc162d90184347e50a890cc8&oe=55DD3400&__gda__=1440572593_c4c4257c8f45c3517dc933a3312f81db

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0First thing you want to do is sketch it out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes u want to or should i?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440417292230:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0srry i dont know how to do it properly its a bit rough

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440417409578:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when it says label coordinates of 4 vertices does that mean what u did on the graph

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440417708934:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0So what is the area you should be integrating

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it where the all curves are

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440418416259:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I guess it wants it above y=8

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0So we probably need a better drawing

phi
 one year ago
Best ResponseYou've already chosen the best response.2the top point is at 9,9^3 or (9,729)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440418815806:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Just putting your pic up in draw phi :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{9}^{2}x ^{3} dx +\int\limits_{9}^{2} 8 dx \int\limits_{9}^{8} x dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it something like this

phi
 one year ago
Best ResponseYou've already chosen the best response.2part (b) wants the integral with respect to y

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0You should have dy

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0So you're doing horizontal strips not vertical, that's part c

phi
 one year ago
Best ResponseYou've already chosen the best response.2the first integral has a right bound formed by y=x the right bound is the curve y= x^3 each rectangle will have width x1x2 and height dy we need x1 and x2 in terms of y x1 (right bound) is x=y so y x2 (left bound is x^3= y so x= y^(1/3) the integral is \[ \int_8^9 yy^\frac{1}{3} \ dy \] then we need a second integral for y=9 to 9^3

phi
 one year ago
Best ResponseYou've already chosen the best response.2can you post the 2nd integral?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{9}^{3} y \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{3}^{9} y dy\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0srry the first one was a mistake

phi
 one year ago
Best ResponseYou've already chosen the best response.2then we need a second integral for y=9 to 9^3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{3}^{9} 8\]

phi
 one year ago
Best ResponseYou've already chosen the best response.2First, identify the dimensions of the "thin rectangle" dw:1440420324066:dw we want the area of that "box" we want a positive number, so we would do Xright  Xleft that gives the width then multiply by the height dy (Xr  Xleft)* dy that is the first step

phi
 one year ago
Best ResponseYou've already chosen the best response.2next step: we don't want to use an x variable. but we know Xright is a constant = 9 so we have (8  Xleft) dy now we need to get rid of Xleft (i.e. put it in terms of y) if we are given a y value, can you find Xleft?

phi
 one year ago
Best ResponseYou've already chosen the best response.2ok, but that x you have in (9x) is some number at a specific y say the "box" is at y=8. what is the value of that x? (remember that x is the x of the point on the curve y=x^3 when y is 8)

phi
 one year ago
Best ResponseYou've already chosen the best response.2yes, in general, (and specifically, when y=8, x is 2) but we want the general answer, because as we change y we want the "correct x" x= y^(1/3) (I would use exponents, easier to integrate)

phi
 one year ago
Best ResponseYou've already chosen the best response.2so fix up this: (9x) dy in other words, replace that x with a function of y

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{3}^{9} [ 9 y ^{\frac{ 1 }{ 3 }} ]\]

phi
 one year ago
Best ResponseYou've already chosen the best response.2that is the area of a single thin box we know "add them up" , which is to say, integrate starting from the lowest y (which is point B in the graph) up to the highest y (which is off the graph)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{8}^{9} yy ^{\frac{ 1 }{ 3}} dy\] +\[\int\limits_{9}^{3} [9y ^{\frac{ 1 }{ 3}}] dy\]

phi
 one year ago
Best ResponseYou've already chosen the best response.2how are you getting 3 for the upper limit?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\int\limits_{3}^{9} [9y ^{\frac{ 1 }{ 3}}] dy

phi
 one year ago
Best ResponseYou've already chosen the best response.2no, the upper limit is where x^3 meets x=9

phi
 one year ago
Best ResponseYou've already chosen the best response.2the left side curve is y= x^3 and the right boundary is x=9 the left side gets closer to x=9 as y gets bigger they intersect when y= 729

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how did u get Y= 729

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0didnt u just use the two integrals and then integrate

phi
 one year ago
Best ResponseYou've already chosen the best response.2the very first part of the problem is to find the 4 points where the various curves meet the shape is like this: dw:1440422110612:dw

phi
 one year ago
Best ResponseYou've already chosen the best response.2we find where y=x^3 meets x=9 obviously they meet when x=9 in the first equation, and the y value is y=9^3 = 729

phi
 one year ago
Best ResponseYou've already chosen the best response.2notice we use two integrals. the first one has a different "right boundary" we use that one from y=8 to y=9 then we switch to the new right boundary (of x=9), and use that all the way up to y=729

phi
 one year ago
Best ResponseYou've already chosen the best response.2so the problem is \[ \int\limits_{8}^{9} yy ^{\frac{ 1 }{ 3}}\ dy + \int\limits_{9}^{729} 9y ^{\frac{ 1 }{ 3}} \ dy\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.06.45+1573.29 =1579.74

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank u so much u have alot of patience

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i show u the solution i have done

phi
 one year ago
Best ResponseYou've already chosen the best response.2part c is more complicated. the first step is draw in a thin vertical rectangle, and figure out its height and width

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440423107863:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so do we get the values respect to x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{2}^{9} x ^{3} dx  \int\limits_{2}^{9} 8 dx  \int\limits_{8}^{9}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{8}^{9} x dx *\]

phi
 one year ago
Best ResponseYou've already chosen the best response.2there are a few ideas we have to keep clear. I would begin with a picture and draw in the rectangle.