Changing cartesian integrals to equivalent polar integrals.

[soccerboy10]

Homework Statement

6...y
∫...∫ x dx dy
0 ...0

The Attempt at a Solution

for reference: π=pi

6...y
∫...∫ x dx dy
0 ...0

y=6=x=6. rcosΘ=6 r=6/cosΘ
goes to

π/2 6secΘ
∫ ...∫... r cosΘ r dr dΘ
π/4...0

π/2.....r=6secΘ
∫ ... [((r^3)/3)cosΘ] dΘ
π/4......r=0

π/2
∫ ...72 sec^2 Θ dΘ
π/4

72tan(π/2) - 72tan(π/4)

last i checked, tan(π/2) was undefined. where did i undergo an error?

 

[Mark44]

Homework Statement

6...y
∫...∫ x dx dy
0 ...0

The Attempt at a Solution

for reference: π=pi

6...y
∫...∫ x dx dy
0 ...0

y=6=x=6. rcosΘ=6 r=6/cosΘ
goes to

π/2 6secΘ
∫ ...∫... r cosΘ r dr dΘ
π/4...0

The upper limit of integration for r is wrong. The region over which integration is taking place is the triangle bounded by the y axis, the line y = x, and the line y = 6. Along this horizontal line, r = 6 cscΘ.

π/2.....r=6secΘ
∫ ... [((r^3)/3)cosΘ] dΘ
π/4......r=0

π/2
∫ ...72 sec^2 Θ dΘ
π/4

72tan(π/2) - 72tan(π/4)

last i checked, tan(π/2) was undefined. where did i undergo an error?

 

[soccerboy10]

how do you integrate csc(cubed) cos?

 

[Mark44]

Please take a look at what I said in my previous post. The integrand is NOT sec³(Θ)cos(Θ)

 

[soccerboy10]

my apologies been a long day studying. problem ended out fine.

 

[Easy_as_Pi]

Not to dig up an old thread, but I have this same problem. I thought this would be better than making an entirely new post. I follow the problem, but am confused as to how the bounds of integration for r were found? I drew the picture, and have the triangle bounded by y=6 at the top, the y axis, and line y=x. I can easily get theta from pi/4 to pi/2, but I'm struggling with finding r. Any advice? Also, this is something I've had issues with on a lot of polar coordinate problems. So, if anyone has a more general explanation as to r, instead of just giving an answer for r's bounds in this problem, that would be greatly appreciated!