Area Interpretation to Evaluate The Integral

In summary,From 0 to 4, the squareroot of (16-x^2) is equal to y and squared to get y^2. The equation is then changed to Y^2 + X^2 = 16. The area of a semi-circle with a radius of 4 is found to be 25.13.
  • #1
swears
87
0
From 0 to 4. ( I don't know how to make the s sign.)

[tex] squareroot of (16-x^2)[/tex]

I set it equal to y and squared it to get.

[tex] y^2 = 16 - x^2[/tex]

Then I changed it to:

[tex] Y^2 + X^2 = 16[/tex]

I know it's a semi circle on the x-axis at 0 to 4.

What should my final answer be?
 
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  • #2
swears said:
From 0 to 4. ( I don't know how to make the s sign.)

[tex] squareroot of (16-x^2)[/tex]

I set it equal to y and squared it to get.

[tex] y^2 = 16 - x^2[/tex]

Then I changed it to:

[tex] Y^2 + X^2 = 16[/tex]

I know it's a semi circle on the x-axis at 0 to 4.

What should my final answer be?


Well do you know how to find the area of a semi-circle with a radius of 4?
 
  • #3
No. But how do you know the radius is 4?

Are ou taking the difference from the x points? Wouldn't the radius be y axis.
 
  • #4
swears said:
No. But how do you know the radius is 4?

Are ou taking the difference from the x points? Wouldn't the radius be y axis.

No? As in you don't know how to find the area of a circle? As to how I know it has radius 4, what is the general equation of a circle?

(x - x0)2 + (y - y0)2 = r2

look at the equation you ahve in your original post and look at teh above, now do you see why it has a radius of 4?
 
  • #5
Ok Thanks.
2 more questions

In the equation of a circle. What do the x subzero's stand for.

And Now I figured out the area to be 25.13 but why do they give me from 0 to 4.
 
  • #6
swears said:
Ok Thanks.
2 more questions

In the equation of a circle. What do the x subzero's stand for.

And Now I figured out the area to be 25.13 but why do they give me from 0 to 4.

The point (x0, y0) would be the center of the circle. They give you from 0 to 4 because they want you to find the integral from 0 to 4 or the area under teh curve from 0 to 4, which in terms of this circle would actually be the area of a quarter circle.
 
  • #7
Ah ok, So do I assume that the center of the circle is the origin here?
 
  • #8
swears said:
Ah ok, So do I assume that the center of the circle is the origin here?

There's no need to ASSUME that it is, because it is and that's all there is to it, compare your equation to the general equation of a circle and the center is obviously the origin.
 
  • #9
Yeah, well my equation is not in your form.

I don't have those x and y subzeroes.
 
  • #10
swears said:
Yeah, well my equation is not in your form.

I don't have those x and y subzeroes.

It more or less is, depending how picky you want to be. The general equation of a circle is as I posted above

(x - x0)2 + (y - y0)2 = r2

So if the center is the origin then x0 = y0 = 0
so the equation becomes

(x - 0)2 + (y - 0)2 = r2

Now note that x - 0 = x and y - 0 = y so now we have

(x)2 + (y)2 = r2

and now removing the parentheses we have

x2 + y2 = r2

Which is the same form as your equation, and note that this will be the general equation for a circle of radius r centered at the origin.
 
  • #11
You do understand, I hope, that you should already have learned what you are talking about- the equation of a circle, and what the numbers in it mean, how to find the area of a circle, or a half or quarter circle, before you start studying calculus!
 
  • #12
It's interesting that you say that because with my school's math program, you do not learn certain important things like this before taking calculus. It's because we have a ridiculously bad integrated program (Math I, Math II, Math III, and Math IV) where all areas of math are covered lightly and gradually throughout high school. Zero rigour. Zero self-teaching (our books only have questions...no answers, examples, etc.).

I guess many kids are going to be very lost next year in my AP Calculus class...hopefully I'll be fine.
 
  • #13
It's hard to believe one would study calculus before basic geometry or even analytic geometry. It looks to me like the whole point of this question is to demonstrate that you can use basic geometry formulas to do integrals. If you don't know the geometry, the whole point is lost!
 
  • #14
Thanks for the help d_leet.
 
  • #15
swears said:
Thanks for the help d_leet.

Your welcome, glad I could help.
 
  • #16
swears said:
Ok Thanks.
2 more questions

In the equation of a circle. What do the x subzero's stand for.

And Now I figured out the area to be 25.13 but why do they give me from 0 to 4.
25.13 is the area of the whole semicircle, which is the integral from -4 to 4. They only want half the semicircle, the half that lies between 0 and 4.
 

FAQ: Area Interpretation to Evaluate The Integral

What is the concept of area interpretation in evaluating integrals?

Area interpretation is a method used in calculus to evaluate the definite integral of a function by interpreting the integral as the area under the curve of the function. This involves breaking up the interval of integration into smaller intervals and approximating the area under the curve using geometric shapes such as rectangles or trapezoids.

How is the Riemann sum used in area interpretation?

The Riemann sum is a mathematical concept used in area interpretation to approximate the area under a curve by dividing the interval of integration into smaller subintervals and finding the sum of the areas of the corresponding rectangles or trapezoids. As the number of subintervals increases, the Riemann sum approaches the exact value of the integral.

What are some common applications of area interpretation in real-life?

Area interpretation is used in various fields such as physics, engineering, and economics to calculate quantities such as velocity, acceleration, and work. It is also used in determining probabilities in statistics and in finding the area under probability density curves in finance and insurance.

How does the concept of area interpretation relate to the fundamental theorem of calculus?

The fundamental theorem of calculus states that the definite integral of a function can be calculated by finding the antiderivative of the function at the limits of integration. This concept is related to area interpretation in that the area under the curve can be calculated by finding the antiderivative of the function and evaluating it at the limits of integration.

Are there any limitations to using area interpretation in evaluating integrals?

One limitation of area interpretation is that it can only be used to evaluate integrals of continuous functions. It also relies on approximations and can be time-consuming for complex functions. In some cases, other methods such as substitution or integration by parts may be more efficient in evaluating integrals.

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