Expert: Paul Klarreich - 5/13/2007

Question
the base of a solid S is the shaded region in the xy-plane enclosed by the x-axis, the y-axis, and the graph of y=1-sinx. for each x, the cross section of S perpendicular to the x-axis at the point (x,O) is an isosceles right triangle whose hypothenuse lies in the xy-plane.
a). find the area of the triangle as a function of x.
b). find the volume of S
thanks

Answer
Questioner:   bhavika
Category:  Calculus
 
Subject:  calculus
Question:  the base of a solid S is the shaded region in the xy-plane enclosed by the x-axis, the y-axis, and the graph of y=1-sinx. for each x, the cross section of S perpendicular to the x-axis at the point (x,O) is an isosceles right triangle whose hypothenuse lies in the xy-plane.
a). find the area of EACH isosceles triangle as a function of x.
b). find the volume of S
thanks
................................
Hi, Bhavika,

Your 'slices' of the volume, S, are those isosceles triangles.  Each

hypotenuse has length  f(x) = 1 - sin x, for  x = 0 to pi/2.  (pi/2 is the

value of x where  f(x) = 0.

If the hypotenuse is  1 - sin x, then each leg is equal to that/sqrt(2).

The area is 1/2 base * height, and we can take each leg as the base or

height.

A = 1/2 [ (1 - sin x)/sqrt(2) ]^2

A = 1/2 (1 - sin x)^2/2

A = 1/4 (1 - sin x)^2

A = 1/4 (1 - 2 sin x + sin^2 x)


That's your area.  Now for the volume, your 'slice' has area A and

thickness dx, so your volume is:

{pi/2
|  1/4 (1 - 2 sin x + sin^2 x) dx
}0

To integrate that, do each term separately:
First term:
{
|  1/4 (1) dx = x/4
}
Second term:
{
|  1/4 (- 2 sin x) dx =  1/2 cos x.  
}
Third term: For this we need the half-angle trick:
{pi/2
|  1/4 (sin^2 x) dx
}0

{pi/2   1 - cos 2x
|  1/4 (----------) dx
}0          2
 x - 1/2 sin 2x
= ---------------
       8


OK, now we do that over the  0..pi/2 interval:
x    cos x     x    sin 2x
--- + ------ + --- - ------ =
4      2       8      16

[pi/8 + 1/2 + pi/16 - 0] - [0 + 1/2 + 0 + 0] =

pi/8 + 1/2 + pi/16 - 1/2
 
pi/8 + pi/16

3pi/16