# Thread: Evaluate the integral geometrically: int[0,a] sqrt{R^2-x^2} dx using triangle, sector

1. ## Evaluate the integral geometrically: int[0,a] sqrt{R^2-x^2} dx using triangle, sector

If someone could explain how to do this problem (in detail please) or show me how, I would greatly appreciate it.

1-2. The function $f(x)\, =\, \sqrt{\strut R^2\, -\, x^2\,}$ has domain $\left[-R,\, R\right].$ Assume $0\, \leq\, a\, \leq\, R.$ We wish to evaluate the following integral:

. . . . .$\displaystyle \int_0^a\, \sqrt{\strut R^2\, -\, x^2\,}\, dx$

(This corresponds to the shaded area in the graphic.)

Tragically, we do not know how to find an antiderivative for $f(x).$ We will learn this later in the course. Instead, evaluate the integral geometrically by splitting the shaded area into the area of a triangle (which is pretty easy) plus the area of a circular sector (which will entail an inverse trig function).

I tried to figure it out in some scribbles on the side.

2. Originally Posted by pleasehelp20
If someone could explain how to do this problem (in detail please) or show me how, I would greatly appreciate it.

1-2. The function $f(x)\, =\, \sqrt{\strut R^2\, -\, x^2\,}$ has domain $\left[-R,\, R\right].$ Assume $0\, \leq\, a\, \leq\, R.$ We wish to evaluate the following integral:

. . . . .$\displaystyle \int_0^a\, \sqrt{\strut R^2\, -\, x^2\,}\, dx$

(This corresponds to the shaded area in the graphic.)

Tragically, we do not know how to find an antiderivative for $f(x).$ We will learn this later in the course. Instead, evaluate the integral geometrically by splitting the shaded area into the area of a triangle (which is pretty easy) plus the area of a circular sector (which will entail an inverse trig function).

I tried to figure it out in some scribbles on the side.
Did you find the two areas suggested? Let's see your result.

3. Originally Posted by tkhunny
Did you find the two areas suggested? Let's see your result.
I have not found either areas. I tried to use the Pythagorean theorem to find the area of the right triangle and the answer did not seem even remotely correct ( I got (a*(R^2-x^2)^1/2)/2) so I erased it. I also tried inverse sin to find the other area but I ended up getting the area R, which also did not seem right.

4. Originally Posted by pleasehelp20
… I tried to use the Pythagorean theorem to find the area of the right triangle and the answer did not seem even remotely correct ( I got (a*(R^2-x^2)^1/2)/2) so I erased it …
First, there is no need to post duplicate threads. Please read the forum guidelines.

Second, the Pythagorean Theorem does not find areas; it relates the three sides (a,b,c) of a right triangle. Symbol c is the length of the hypotenuse, and symbols a and b are the lengths of the legs.

a^2 + b^2 = c^2

The area of a right triangle is found by formula:

Area = 1/2 · base · height

Can you see that the height is f(a)?

So, you have expressions for the base and height (in terms of R and a). Plug them into the area formula.

5. I would recommend using limits and sums to find the area under the curve.

This should help
http://goblues.org/faculty/kollathl/...Rectangles.pdf

6. Originally Posted by pleasehelp20
… I got (a*(R^2-x^2)^1/2)/2…
Oh, this is so close.

I ought to have looked more carefully, last night. I think your statement about using Pythagoras for finding area threw me. Here's your result, rearranged just a bit, to eliminate those outer grouping symbols:

Area of right triangle = a/2*(R^2 - x^2)^(1/2)

Change one symbol, and it's correct! Again, the height of the triangle is f(a). You've used f(x) instead of f(a).

7. Originally Posted by chris84567
I would recommend using limits and sums to find the [shaded] area …
In this exercise, the instructions state the method that is to be used, and it's a geometrical approach dealing with the indicated right triangle and circular sector.

There's another geometrical approach (which I think is easier than dealing with the sector). That method takes half the difference between two circular segments.

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