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

Click For Summary

Homework Help Overview

The problem involves finding a horizontal line y=k that divides the area between the curves y=x^2 and y=9 into two equal parts. The context is rooted in integral calculus and area calculations under curves.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the total area between the curves and how to set up integrals to find the value of k. There are attempts to clarify the limits of integration and the reasoning behind the area calculations. Some participants express confusion about the area values and the setup of the integrals.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the setup of the integrals and the interpretation of the area under the curves. Some guidance has been provided regarding the integral setup, but there is no consensus on the correct approach or final value of k.

Contextual Notes

Participants question the assumptions regarding the area calculations and the intercepts of the curves. There is mention of potential typos in the integral expressions, and confusion persists about the total area and how it relates to the value of k.

SAT2400
Messages
68
Reaction score
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
 
Physics news on Phys.org
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
 
Last edited by a moderator:


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
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
Replies
3
Views
2K
Replies
12
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
12
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
11
Views
2K
Replies
6
Views
2K
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
5K