Results 1 to 2 of 2

Math Help - Area of Archway

  1. #1
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432

    Area of Archway

    I have this really hard question which is attached at the bottom. I am not supposed to use a calculator to solve it but I've been trying a few methods to work it out, and nothing has come nearer than calculus. I used the equation of a circle graph with radius 2 and found the integral from R to P, and multiplied it by 2.

    However, I haven't actually technically 'learnt' calculus so I had to solve it with my calculator and got this:

    Area from P to R = \int_{-2}^{-1}\sqrt{4-x^2}dx=\frac{4sin^{-1}(\frac{x}{2})}{90}+\frac{x\sqrt{4-x^2}}{2}

    Some devilish answer... and then there's all the tricky math involved in evaluating it. Anyways, after running it through the calculator and multiplying by 2 I get (B).

    Is there any simple (non-calculus?) way to solve this? It was a question on a very time-limited test. Thanks.
    Attached Thumbnails Attached Thumbnails Area of Archway-pqr.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by DivideBy0 View Post
    I have this really hard question which is attached at the bottom. I am not supposed to use a calculator to solve it but I've been trying a few methods to work it out, and nothing has come nearer than calculus. I used the equation of a circle graph with radius 2 and found the integral from R to P, and multiplied it by 2.

    However, I haven't actually technically 'learnt' calculus so I had to solve it with my calculator and got this:

    Area from P to R = \int_{-2}^{-1}\sqrt{4-x^2}dx=\frac{4sin^{-1}(\frac{x}{2})}{90}+\frac{x\sqrt{4-x^2}}{2}

    Some devilish answer... and then there's all the tricky math involved in evaluating it. Anyways, after running it through the calculator and multiplying by 2 I get (B).

    Is there any simple (non-calculus?) way to solve this? It was a question on a very time-limited test. Thanks.

    the area will be twice the area of sector PQR minus the area of the equilateral triagle formed by their radii. (see diagram below)

    so A = 2*area of sector in radians - 2*(1/2)bh

    so A = 2( (1/2)r^2(pi/3)) - 2((1/2)(1)(sqrt(3))
    => A = 2(2pi/3) - sqrt(3)
    => A = 4pi/3 - sqrt(3)

    any questions?
    Attached Thumbnails Attached Thumbnails Area of Archway-arc.gif  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. area-preserving maps between regions of equal area
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: August 23rd 2010, 08:33 PM
  2. Replies: 3
    Last Post: May 25th 2010, 12:01 AM
  3. Lateral Area and Total Surface Area
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 25th 2009, 04:28 PM
  4. Replies: 2
    Last Post: November 10th 2008, 09:27 AM
  5. Replies: 4
    Last Post: April 20th 2008, 07:03 PM

Search Tags


/mathhelpforum @mathhelpforum