anonymous
  • anonymous
hello, advance definite integral area bound by 4 curves. https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-xpl1/v/t34.0-12/11939020_821736254611990_2090835384_n.jpg?oh=3cd85831dc162d90184347e50a890cc8&oe=55DD3400&__gda__=1440572593_c4c4257c8f45c3517dc933a3312f81db
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@King.Void.
Astrophysics
  • Astrophysics
First thing you want to do is sketch it out
anonymous
  • anonymous
yes u want to or should i?

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Astrophysics
  • Astrophysics
You should
anonymous
  • anonymous
|dw:1440417292230:dw|
anonymous
  • anonymous
srry i dont know how to do it properly its a bit rough
Astrophysics
  • Astrophysics
|dw:1440417409578:dw|
anonymous
  • anonymous
yes
anonymous
  • anonymous
when it says label co-ordinates of 4 vertices does that mean what u did on the graph
anonymous
  • anonymous
|dw:1440417708934:dw|
Astrophysics
  • Astrophysics
So what is the area you should be integrating
anonymous
  • anonymous
is it where the all curves are
anonymous
  • anonymous
|dw:1440418416259:dw|
Astrophysics
  • Astrophysics
I guess it wants it above y=8
Astrophysics
  • Astrophysics
So we probably need a better drawing
phi
  • phi
it wants this region
1 Attachment
Astrophysics
  • Astrophysics
Yeah exactly
phi
  • phi
the top point is at 9,9^3 or (9,729)
Astrophysics
  • Astrophysics
|dw:1440418815806:dw|
Astrophysics
  • Astrophysics
Just putting your pic up in draw phi :P
anonymous
  • anonymous
\[\int\limits_{9}^{2}x ^{3} dx +\int\limits_{9}^{2} 8 dx -\int\limits_{9}^{8} x dx\]
anonymous
  • anonymous
is it something like this
phi
  • phi
part (b) wants the integral with respect to y
anonymous
  • anonymous
yes
Astrophysics
  • Astrophysics
You should have dy
Astrophysics
  • Astrophysics
So you're doing horizontal strips not vertical, that's part c
anonymous
  • anonymous
how?
phi
  • phi
|dw:1440419331010:dw|
phi
  • phi
the first integral has a right bound formed by y=x the right bound is the curve y= x^3 each rectangle will have width x1-x2 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 \[ \int_8^9 y-y^\frac{1}{3} \ dy \] then we need a second integral for y=9 to 9^3
phi
  • phi
follow?
anonymous
  • anonymous
yes
phi
  • phi
can you post the 2nd integral?
anonymous
  • anonymous
\[\int\limits_{9}^{3} y \]
anonymous
  • anonymous
\[\int\limits_{3}^{9} y dy\]
anonymous
  • anonymous
srry the first one was a mistake
phi
  • phi
then we need a second integral for y=9 to 9^3
anonymous
  • anonymous
\[\int\limits_{3}^{9} 8\]
phi
  • phi
First, 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
anonymous
  • anonymous
ok
phi
  • phi
next 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
  • phi
**(9 - Xleft) dy
anonymous
  • anonymous
(9-x) dy
phi
  • phi
ok, but that x you have in (9-x) 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)
anonymous
  • anonymous
\[\sqrt[3]{y}\]
phi
  • phi
yes, 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
  • phi
so fix up this: (9-x) dy in other words, replace that x with a function of y
anonymous
  • anonymous
(9-y^(1/3) ) dy
anonymous
  • anonymous
\[\int\limits_{3}^{9} [ 9- y ^{\frac{ 1 }{ 3 }} ]\]
anonymous
  • anonymous
dy
phi
  • phi
that 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
  • anonymous
ok
anonymous
  • anonymous
\[\int\limits_{8}^{9} y-y ^{\frac{ 1 }{ 3}} dy\] +\[\int\limits_{9}^{3} [9-y ^{\frac{ 1 }{ 3}}] dy\]
phi
  • phi
how are you getting 3 for the upper limit?
anonymous
  • anonymous
srry mistake
anonymous
  • anonymous
\int\limits_{3}^{9} [9-y ^{\frac{ 1 }{ 3}}] dy
phi
  • phi
no, the upper limit is where x^3 meets x=9
phi
  • phi
the 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
  • anonymous
how did u get Y= 729
anonymous
  • anonymous
didnt u just use the two integrals and then integrate
phi
  • phi
the 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
  • phi
we 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
  • phi
notice 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
anonymous
  • anonymous
yea
phi
  • phi
so the problem is \[ \int\limits_{8}^{9} y-y ^{\frac{ 1 }{ 3}}\ dy + \int\limits_{9}^{729} 9-y ^{\frac{ 1 }{ 3}} \ dy\]
anonymous
  • anonymous
oh wow
anonymous
  • anonymous
6.45+1573.29 =1579.74
anonymous
  • anonymous
is this right?
phi
  • phi
yes
anonymous
  • anonymous
thank u so much u have alot of patience
anonymous
  • anonymous
part c?
anonymous
  • anonymous
i show u the solution i have done
phi
  • phi
part c is more complicated. the first step is draw in a thin vertical rectangle, and figure out its height and width
anonymous
  • anonymous
|dw:1440423107863:dw|
anonymous
  • anonymous
so do we get the values respect to x
anonymous
  • anonymous
\[\int\limits_{2}^{9} x ^{3} dx - \int\limits_{2}^{9} 8 dx - \int\limits_{8}^{9}\]
anonymous
  • anonymous
\[\int\limits_{8}^{9} x dx *\]
phi
  • phi
there are a few ideas we have to keep clear. I would begin with a picture and draw in the rectangle.
phi
  • phi
|dw:1440423583733:dw|