I was instructed to do this problem using polar coordinates

I was instructed to do this problem using polar coordinates. Not sure if x^2 + y^2 in the function can just be substituted with r^2 and how that would work in the 1 = double integral. Any help would be appreciated.

If f(x,y)=c(x2+y2)3/2 for 1 le x2+y2 le 25 and x ge 0,then find: the value of c to make this density function. P(x2+y2>16). P(x>2y).sets up only.

Solution

I would like you to try the problem by taking some good hints from below,Ping me again if you want a detailed solution,i will edit my answer

First look at the conditions given

as you guessed righty x^2+y^2 can be replaced as r and at the same time look at the other condition x>=0

which can be replaced as rcosa>=0

it means cosa>=0 that means the region of integration is from -pi/2 to pi/2

so while changing to polar also see that the region represents a circle of radius 1 and 5 in polar

as r^2=1,25 The region between r=1 and r=5 and a=(-pi/2 to pi/2)

Now as you know the limits just plug in conditions

as fxy is a density function ,so total probability must be 1 integrate fxy using the limits i told above and put it equal to 1.You will get the value of c

now use that value of c for B) and C) parts

for B) part you just have to change the integration limits integrate fxy(in polar replace x by rcosa and y be rsina)

from the limits r=4 to r=5 as the B) part region is x^2+y^2>16 which is the area between r=4 and r=5 in polar

for C) part use rcosa>rsina which is equivalent to cota>1 as your original region of function is r=1 to r=5 and angle in 1st and 4th quadrant and now condition given is cota>1 the intersection of region is 1st quadrant.

for integration of area use 0.5 r^2 da where a is small element of angle and the limits are angle itself

Hope this helps..