00:01
Hi, in this question, the formula for the polar radius of the gyration is k not is equal to square root of i by a.
00:06
I is the moment of inertia of the area about the polar axis.
00:09
In this case, the x axis and a is the area of the shaded region.
00:12
To find, we can use the formula that i is equal to integral r square t a.
00:19
So, where r is the distance from the polar axis to each element of the area da.
00:23
Since the shaded region is symmetric about the y axis, we can split it into two equal ors and integrate from y is equal to 0 to y is equal to b by 2, where b is the height of the shaded region.
00:33
So, for each of we can express r in terms of y as r is equal to a plus b by 2 minus y.
00:41
The area element is that da is equal to 2 dx dy, where dx is small element of the width along the x axis.
00:50
Now, we can express dx in terms of dy as dx is equal to 2a into b by 2 minus y by b.
00:59
Substituting r and dx into the formula for i, we get that i is equal to 2 into integral of a plus b by 2 minus y the whole square 2a b by 2 minus y by b dy.
01:20
So, which is equal to 8a cube by 3 plus 2ab square by 3.
01:32
Now, to find a, we can integrate the equation of the curve that forms the boundary of the shaded region.
01:38
So, from the diagram, we can see that this curve is given by that r is equal to a plus b by 2 minus y and this implies that r minus a is equal to b by 2 minus y.
01:52
So, this implies y is equal to b by 2 minus r minus a.
01:58
So, this is semicircle with radius a, center at a comma b by 2.
02:02
The equation of the circle is as we all know that x minus a the whole square plus y minus b the whole square y minus b by 2 the whole square which is equal to a square.
02:23
Now, substitute y is equal to b by 2 minus r minus a, we get that x minus a the whole square plus b by 2 minus r minus a the whole square is equal to a square.
02:41
This implies that r square minus 2a r plus a square plus b by 2 the whole square minus b of r minus a is equal to 0.
02:53
Now, solving for r, we get r is equal to a plus b by 2 plus or minus root of a square minus b by 2 the whole square plus b into r minus a.
03:10
So, which is since r is always greater than or equal to a, we take the positive root...