Results 1 to 7 of 7

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

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

    . . . . .[tex]\displaystyle \int_0^a\, \sqrt{\strut R^2\, -\, x^2\,}\, dx[/tex]

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

    Tragically, we do not know how to find an antiderivative for [tex]f(x).[/tex] 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.
    Attached Images Attached Images
    Last edited by stapel; 01-25-2018 at 06:55 PM. Reason: Typing out the text in the graphic; creating useful subject line.

  2. #2
    Elite Member
    Join Date
    Apr 2005
    Location
    USA
    Posts
    9,283
    Quote Originally Posted by pleasehelp20 View Post
    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 [tex]f(x)\, =\, \sqrt{\strut R^2\, -\, x^2\,}[/tex] has domain [tex]\left[-R,\, R\right].[/tex] Assume [tex]0\, \leq\, a\, \leq\, R.[/tex] We wish to evaluate the following integral:

    . . . . .[tex]\displaystyle \int_0^a\, \sqrt{\strut R^2\, -\, x^2\,}\, dx[/tex]

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

    Tragically, we do not know how to find an antiderivative for [tex]f(x).[/tex] 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.
    Last edited by stapel; 01-25-2018 at 06:55 PM. Reason: Copying typed-out graphical content into reply.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  3. #3
    Quote Originally Posted by tkhunny View Post
    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. #4
    Elite Member mmm4444bot's Avatar
    Join Date
    Oct 2005
    Location
    Seattle
    Posts
    9,347
    Quote Originally Posted by pleasehelp20 View Post
    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.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  5. #5
    New Member chris84567's Avatar
    Join Date
    Jan 2018
    Location
    A Concave Earth
    Posts
    12
    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. #6
    Elite Member mmm4444bot's Avatar
    Join Date
    Oct 2005
    Location
    Seattle
    Posts
    9,347
    Quote Originally Posted by pleasehelp20 View Post
    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).
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  7. #7
    Elite Member mmm4444bot's Avatar
    Join Date
    Oct 2005
    Location
    Seattle
    Posts
    9,347
    Quote Originally Posted by chris84567 View Post
    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.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

Tags for this Thread

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •