Question

PLEASE HELP!!!! I have somehow forgot how to find the place where two graphs intersect. Question: "Find the solutions to the graphs y=x^2 , and x=y^2"

Answer 1

y=x^2 and x=y^2
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You want to find common intersection points.

1) y = x^2
2) x = y^2


Solve by substitution.

Take equation 1) y = x^2 and in place of x, substitute from equation 2) y^2 for x.

Answer 2

That gives 1) y = (x)^2 = (y^2)^2 = y^4.

y = y^4
y^4 = y
Collect ys on on one side and factor

y^4 - y = 0

y (y^3 - 1) = 0

Zero Product Property

y = 0 or y^3 - 1 = 0

Answer 3

For the y = 0, find the x that goes with it.
That is one point of intersection.

For the next point of intersection, factor y^3 - 1 = 0.

I am attaching the pattern for factoring the sum/difference of two cube.

@chrisplusian

Answer 4

Wow I feel rusty I learned this about five semesters ago and now I couldn't remember, it seems so trivial and Now I remember thank you

Answer 5

couldn't you solve simply by adding one and then taking the cubed root of both sides to then find that y=1?

Answer 6

You are welcome.

Post what you get for the points of intersection.

You can also see them on the graphing program here:

http://www.wolframalpha.com/input/?i=y%3Dx%5E2%2C++x%3Dy%5E2

Answer 7

>>couldn't you solve simply by adding one and then taking the cubed root of both sides to then find that y=1?

Try it and see it that works.

Answer 8

It seems as though it does. Doing it that way gives me y=1, and y=0 and plugging that back into both equations gives the same point so I think it does. do you agree?

Answer 9

The points of intersection have coordinates (0,0) and (1,1).

y^3 -1 = 0 has three solutions. One of them is 1. The other two are not real numbers. Because I did not know over what set of numbers you wanted to solve, I went with the difference of two cubes factoring pattern. A good refresher, if nothing else.

Answer 10

I see, I was trying to calculate the center of mass of a lamina, and just needed to find my bounds of integration over the region, so I only need to be concerned with the real, and not complex numbers so that works, thanks again

Answer 11

You are welcome.

Answer 12

@Directrix can I ask you one other conceptual question? What am I doing when I take a triple integral? I know a double integral of a function is finding volume:\[\int\limits_{a}^{b}\int\limits_{c}^{d}f(x,y)da=volume\]but what is a triple integral of a function? ex:\[\int\limits_{a}^{b}\int\limits_{c}^{d}\int\limits_{e}^{f}f(x,y,z)dv\]

Answer 13

We used a double integral to integrate over a two-dimensional region and so it shouldn’t be too surprising that we’ll use a triple integral to integrate over a three dimensional region.
http://tutorial.math.lamar.edu/Classes/CalcIII/TripleIntegrals.aspx

http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/255trip/255trip.html

Answer 14

@Directrix so to summarize this a double integral over a function of two variables is the same as volume. But this is true because the lower bound of integration is z=0. However one interpretation of a triple integral over a 3d region is actually volume but it is a tripe integral because it may not have z=0 as one of the bounds of integration? I hope this doesn't sound silly I am just trying to figure it all out

Answer 15

Or is it more logical to say that it doesn't have a geometrical interpretation? Because if we integrate a single variable function then we get area, if we integrate a function of to variables twice (double integration) we get volume, then would a triple integral of a function in three variables represent something in the fourth dimension?

Answer 16

Someone who is more into Calculus than I am needs to respond. I'll post the question and route it here.

Answer 17

\[\int\limits \int\limits \int\limits dV = V\]

If you're integrating a 3-space function over 3-space then you get a 4 space solution. It may have practical implications even though the geometry seems a bit wacky. You could integrate a density function over 3 space to get the total mass of an object. nothing wacky about that, yeah?

Answer 18

makes sense thanks for all the help

Answer 19

@chrisplusian See the response from @Peter14
Click on attachment.