Ind a horizontal line y=k that divides the area* QUESTION*

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SAT2400
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ind a horizontal line y=k that divides the area*URGENT QUESTION*

Homework Statement


Find a horizontal line y=k that divides the area between y=x^2 and y=9into two equal parts.


Homework Equations


integral b,a, ( f(x) -g(x) ) dx


The Attempt at a Solution



half of the total area: 18
Integral(b,a) (k -x^2) dx = 18
Integral(b,a) (9-k) dx = 18


Please explain how to solve this ! T_T
 
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SAT2400 said:

Homework Statement


Find a horizontal line y=k that divides the area between y=x^2 and y=9into two equal parts.


Homework Equations


integral b,a, ( f(x) -g(x) ) dx


The Attempt at a Solution



half of the total area: 18
Integral(b,a) (k -x^2) dx = 18
Integral(b,a) (9-k) dx = 18


Please explain how to solve this ! T_T

Where does y=k intercept x^2? There is a couple of solutions; set 'em as the limits of the integral [tex]\int(9-x^2)dx=18[/tex]. This is a simple cubic equation that has certainly one real solution. (To solve it use Cardano's method and http://www.trans4mind.com/personal_development/mathematics/polynomials/cardanoMethodExamples.htm"you can find the solved examples of how to do it.)

AB
 
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Can you explain more in detail?

why do you do integral(b,a) (9- x^2) dx = 18?? isn't it 36?

y=9 is the top part

y=k will be the middle part of two areas(each 18)

y=x^2 is the bottom part..?

Please help! T_T very desperate
 


SAT2400 said:
Can you explain more in detail?

why do you do integral(b,a) (9- x^2) dx = 18?? isn't it 36?

y=9 is the top part

y=k will be the middle part of two areas(each 18)

y=x^2 is the bottom part..?

Please help! T_T very desperate

Just graph the function y=x^2 and see where y=9 intercepts it. The area surrounded by x^2and y=9 is 36; so the half of it will be encircled by y=k which intercepts x^2 at [tex]\pm \sqrt{k}[/tex].

And one typo: I must have typed [tex]\int(k-x^2)dx=18[/tex]. (Sorry for inconvenience)

AB
 


how do i find the value of K??!?~

(-unknown x , k) and (unknown x , k) will be the interceptions btwn y=k and y=x^2

But/
still confused :(
 


SAT2400 said:
how do i find the value of K??!?~

(-unknown x , k) and (unknown x , k) will be the interceptions btwn y=k and y=x^2

But/
still confused :(

Is solving [tex]\int^{+\sqrt{k}}_{-\sqrt{k}}(k-x^2)dx=18[/tex] the problem you have?

Here we go: [tex]\int^{+\sqrt{k}}_{-\sqrt{k}}(k-x^2)dx=|kx-\frac{1}{3}x^3|^{x=+\sqrt{k}}_{x=-\sqrt{k}}=|2k^{3/2}-\frac{2}{3}k^{3/2}|=18[/tex]. Now continuing this is your job.

AB