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

- anonymous

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- schrodinger

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- anonymous

@King.Void.

- Astrophysics

First thing you want to do is sketch it out

- anonymous

yes u want to or should i?

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## More answers

- Astrophysics

You should

- anonymous

|dw:1440417292230:dw|

- anonymous

srry i dont know how to do it properly its a bit rough

- Astrophysics

|dw:1440417409578:dw|

- anonymous

yes

- anonymous

when it says label co-ordinates of 4 vertices does that mean what u did on the graph

- anonymous

|dw:1440417708934:dw|

- Astrophysics

So what is the area you should be integrating

- anonymous

is it where the all curves are

- anonymous

|dw:1440418416259:dw|

- Astrophysics

I guess it wants it above y=8

- Astrophysics

So we probably need a better drawing

- phi

it wants this region

##### 1 Attachment

- Astrophysics

Yeah exactly

- phi

the top point is at 9,9^3 or (9,729)

- Astrophysics

|dw:1440418815806:dw|

- Astrophysics

Just putting your pic up in draw phi :P

- anonymous

\[\int\limits_{9}^{2}x ^{3} dx +\int\limits_{9}^{2} 8 dx -\int\limits_{9}^{8} x dx\]

- anonymous

is it something like this

- phi

part (b) wants the integral with respect to y

- anonymous

yes

- Astrophysics

You should have dy

- Astrophysics

So you're doing horizontal strips not vertical, that's part c

- anonymous

how?

- phi

|dw:1440419331010:dw|

- 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

follow?

- anonymous

yes

- phi

can you post the 2nd integral?

- anonymous

\[\int\limits_{9}^{3} y \]

- anonymous

\[\int\limits_{3}^{9} y dy\]

- anonymous

srry the first one was a mistake

- phi

then we need a second integral for y=9 to 9^3

- anonymous

\[\int\limits_{3}^{9} 8\]

- 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

ok

- 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

**(9 - Xleft) dy

- anonymous

(9-x) dy

- 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

\[\sqrt[3]{y}\]

- 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

so fix up this:
(9-x) dy
in other words, replace that x with a function of y

- anonymous

(9-y^(1/3) ) dy

- anonymous

\[\int\limits_{3}^{9} [ 9- y ^{\frac{ 1 }{ 3 }} ]\]

- anonymous

dy

- 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

ok

- anonymous

\[\int\limits_{8}^{9} y-y ^{\frac{ 1 }{ 3}} dy\] +\[\int\limits_{9}^{3} [9-y ^{\frac{ 1 }{ 3}}] dy\]

- phi

how are you getting 3 for the upper limit?

- anonymous

srry mistake

- anonymous

\int\limits_{3}^{9} [9-y ^{\frac{ 1 }{ 3}}] dy

- phi

no, the upper limit is where x^3 meets x=9

- 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

how did u get Y= 729

- anonymous

didnt u just use the two integrals and then integrate

- 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

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

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

yea

- 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

oh wow

- anonymous

6.45+1573.29 =1579.74

- anonymous

is this right?

- phi

yes

- anonymous

thank u so much u have alot of patience

- anonymous

part c?

- anonymous

i show u the solution i have done

- phi

part c is more complicated.
the first step is draw in a thin vertical rectangle, and figure out its height and width

- anonymous

|dw:1440423107863:dw|

- anonymous

so do we get the values respect to x

- anonymous

\[\int\limits_{2}^{9} x ^{3} dx - \int\limits_{2}^{9} 8 dx - \int\limits_{8}^{9}\]

- anonymous

\[\int\limits_{8}^{9} x dx *\]

- phi

there are a few ideas we have to keep clear.
I would begin with a picture
and draw in the rectangle.

- phi

|dw:1440423583733:dw|