?Evaluating Double Integral Using Polar Coords

[joemama69]

Homework Statement

Use polar coords to evaluate the double integral x³ + xy²dydx from y = -(9-x²)^1/2 to (9-x²)^1/2, and x = 0 to 3

Homework Equations

The Attempt at a Solution

So the region is a half circle of radius 3, centered @ the origin, with only the possitive x side, (right side of circle)

x = rcosQ, y = rsinQ

the integral is r³cos³Q + r³cosQsinQ drdQ

after integration with respect to r i get and pluging in the limits

84/4 cos^3 Q + 81/4 cosQsinQ dQ

i looked up the integrals of cos^3Q in the back of my book and it is quite complex. Have i made any mistakes thus far, and is there an easier way to evaluate it

 

[dx]

x³ + xy² = r² cosθ. Check your algebra.

 

[Cyosis]

Also if you go from Cartesian to polar coordinates dxdy \Rightarrow r dr d\theta.
While not relevant to this problem \cos^3x is easily integrated by writing it as \cos x (1-\sin^2x) and substituting u=\sin x.

 

[joemama69]

i think i messed up somehwere

\int r^3cosQdrdQ = .25r^4cosQ from 0 to 3
=\int81/4 cosQ dQ from 0 to pi = 81/4 sinQ from 0 to pi
sin(0) = 0, sin (pi) = 0
i got an answer of 0

 

[dx]

Are you sure θ goes from 0 to π?

 

[Cyosis]

You forgot that when you transform from Cartesian to polar coordinates you have to add the Jacobian r (dx dy \Rightarrow r dr d\theta). Secondly your limits for \theta are off. You are integrating over the upper semi circle, but that is not the right region to integrate over. Draw the circle and add the Cartesian limits, over what section should you be integrating?

 

[dx]

Q should go from -\frac{\pi}{2} to \frac{\pi}{2}.

 

[HallsofIvy]

You forgot that when you transform from Cartesian to polar coordinates you have to add the Jacobian r (dx dy \Rightarrow r dr d\theta). Secondly your limits for \theta are off. You are integrating over the upper semi circle, but that is not the right region to integrate over. Draw the circle and add the Cartesian limits, over what section should you be integrating?

Actually, he did include the "r" but you are correct that his limits of integration on \theta are wrong.

 

[Cyosis]

You're right I completely missed the r.

 

[joemama69]

uhhh huh, ok i didnt realize that mattered, i figured as long as it was 180 degrees it would be the same

but i think i did mess up the r

x(x² + y²) = rcosQ* r² = r³cosQ


then i add the jacobian r and my integral is for r⁴cosQdrdQ

the only thing that changes is the 81/4 becomes 243/5

so the final answer is

243/5 + 243/5 = 486/5

 

[Cyosis]

That's correct.