Lower Level integration problem (Find the centroid)

In summary, the student attempted to find the mass of a region bounded by the x-axis and \sqrt{9-x^2}, but ran into difficulty integrating the equation. Once he substituted x=3sin(t) into the equation, the answer came by itself.
  • #1
J-villa
4
0
EDIT - Solved. Thank you.

Homework Statement


Find the Centroid of the region bounded by the x-axis and [tex]\sqrt{9-x^2}[/tex]

Homework Equations


So far, I found my Mx which ended up being 27p after applying this forumula:
p[tex]\int[/tex]([tex]\frac{\sqrt{9-x^2}+0}{2}[/tex]) ([tex]\sqrt{9-x^2}[/tex]-0)dx​
*My interval was [-3,3]

The Attempt at a Solution



My problem is when I go to find the mass so I can find the centroid and get my final answer. The equation to find the mass is:

m=p[tex]\int[/tex][f(x)-g(x)]dx

so when applying the formula to my problem, I get this:

m=p[tex]\int[/tex][tex]\sqrt{9-x^2}[/tex]

So really, it ends up being a simple integration stumble. I just can't figure out how to integrate this. I tried the substitution rule, but it didn't work out.

Can someone point me in the right direction please?
 
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  • #2
I'm guessing you have probably gone wrong somewhere unfortunately i don't know where. But the solution to your integral:

http://www5a.wolframalpha.com/Calculate/MSP/MSP8641961aa713bih43db00000ebbd4gc0a51fd4i?MSPStoreType=image/gif&s=15
 
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  • #3
Thank you greatly for the reply. I'm in Calc. II, learning all this integration as fast as I can and I stumble here and there in my theorems.

I was looking at my inverse and hyperbolic trig. functions to try and make a connection, but never thought of double substituting regular sine.

However, I'm not the kinda person to write an answer/work down and claim it as my own, especially since I won't be able to do it when the test comes, so using your reply I have a few questions.
  • #1 - I notice you didn't see the constant (p) I have to have prior to the integral. I'm guessing that the p would just go along for the ride with the all the other constants.
  • #2 - speaking of constants, how did you get that 9 outside the integral (Step:1/2)? I was thinking you just pulled the 9 out from the integrand; but wouldn't that leave a 1-sin^2?
Once again, thanks for the responses. I'm going to be lurking on these forums a lot more often now.
 
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  • #4
We are using the constant multiple rule for integration and 1-sin^2=cos^2.
 
  • #5
elliotician said:
We are using the constant multiple rule for integration and 1-sin^2=cos^2.

oh, duh'. I appreciate the help. I'll respond soon w/ my answer.
 
  • #6
substitute x=3sin(t)..and the answer comes by itself...
 

FAQ: Lower Level integration problem (Find the centroid)

What is the Lower Level integration problem?

The Lower Level integration problem is a mathematical concept that involves finding the centroid of a given shape or region. The centroid is the point at which the shape would balance if it were cut out of a sheet of material. It is also known as the geometric center or center of mass.

Why is finding the centroid important?

Finding the centroid is important in many fields, such as engineering, physics, and architecture. It allows us to determine the stability and balance of a structure, as well as its moments of inertia and other important properties. It also helps in creating efficient designs and calculations for various applications.

What is the difference between lower level and higher level integration problems?

Lower level integration problems involve finding the centroid of a two-dimensional shape or region, while higher level integration problems involve finding the centroid of a three-dimensional object. Higher level integration problems also require more complex mathematical techniques, such as triple integrals, compared to the simpler double integrals used in lower level problems.

How is the centroid calculated in lower level integration problems?

The centroid of a shape can be calculated by dividing the integral of each coordinate (x and y) by the total area of the shape. The integral of each coordinate is calculated by multiplying the coordinate by the function representing the shape, then integrating over the region. The total area of the shape is calculated by integrating the function over the region as well.

What are some real-world applications of the lower level integration problem?

The lower level integration problem has many real-world applications, such as finding the center of mass for a physical object, determining the balance and stability of structures, calculating moments of inertia for rotating objects, and designing efficient shapes for various purposes, such as airplane wings and car bodies. It also has applications in computer graphics, robotics, and computer-aided design (CAD).

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