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

In summary: Here we go: \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. Now continuing this is your job.
  • #1
SAT2400
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0
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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  • #2
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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  • #3


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
 
  • #4


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
 
  • #5


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 :(
 
  • #6


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
 

1. How do I find the value of k for a horizontal line that divides the area?

The value of k for a horizontal line that divides the area can be found by taking the average of the highest and lowest y-values in the given area. This will give you the value of k, which will divide the area into two equal parts.

2. How do I graph a horizontal line y=k that divides the area?

To graph a horizontal line y=k that divides the area, plot the point (0,k) on the coordinate plane. Then, draw the line horizontally across the plane. This line will divide the area into two equal parts.

3. Is it possible for a horizontal line to divide the area into unequal parts?

No, a horizontal line will always divide the area into two equal parts. This is because a horizontal line has a constant y-value, meaning that it will always intersect the x-axis at the same point, dividing the area into equal parts.

4. How can I use the equation y=k to find the horizontal line that divides the area?

The equation y=k represents a horizontal line with a y-intercept of k. By setting k to the average of the highest and lowest y-values in the given area, you can find the equation for the horizontal line that divides the area into two equal parts.

5. Can I use a horizontal line to divide the area of any shape?

Yes, a horizontal line can be used to divide the area of any shape. As long as the line is drawn horizontally and intersects the x-axis at the average of the highest and lowest y-values, it will divide the area into two equal parts.

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