Alamina with constant density rho(x, y) = rho occupies the given region. find the moments of inertia ix and iy and the radii of gyration x double bar and y double bar. the part of the disk x2 + y2 ≤ a2 in the first quadrant

Ix = Iy =

Radius of gyration x = y =  

Step-by-step explanation:

Given: A lamina with constant density ρ(x, y) = ρ occupies the given region x2 + y2 ≤ a2 in the first quadrant.

Mass of disk = ρπR2

Moment of inertia about its perpendicular axis is . Moment of inertia of quarter disk about its perpendicular is .

Now using perpendicular axis theorem, Ix = Iy = = .

For Radius of gyration K, equate MK2 = MR2/16, K= R/4.

the answer is a i belaeve atleast thats what my calculator just gfave me

step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.