Evaluate the iterated integral by converting to polar coordinates

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Homework Statement



Where the region is:

D = {(x,y)| 0[tex]\leq[/tex]x[tex]\leq[/tex]2;0[tex]\leq[/tex]y[tex]\leq[/tex][tex]\sqrt{}2x-x^2[/tex]}

Double integral over region D with f(x,y) = [tex]\sqrt{}x^2+y^2[/tex] and respect to dA

Homework Equations


Trig. Identities:
x = rcos(theta)
y = rsin(theta)
x^2+y^2 = r^2


The Attempt at a Solution



First, I graphed D in xy-plane and got a semi-circle with its center translated to the right by 1 units and is bounded by
x=0 and x=2
I converted to polar coordinates and got these:
f(x,y) = r
dA = r(dr)d(theta)

D = {(r,theta)| 0[tex]\leq[/tex]r[tex]\leq[/tex]2;x/0[tex]\leq[/tex]theta[tex]\leq[/tex]pi}

I'm not sure if everything is accounted from (x,y) to (r,theta) and the cone that is made from f(x,y).

Thank you for taking your time to read this.
 

Answers and Replies

  • #2
SammyS
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What is the equation of the "(semi-)circle with its center translated to the right by 1 units" (radius = 1) ? - in polar coordinates.
 
  • #3
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What is the equation of the "(semi-)circle with its center translated to the right by 1 units" (radius = 1) ? - in polar coordinates.

it's the function of the y interval --> [tex]\sqrt{}2x-x^2[/tex]
I didn't convert it to polar coordinates, I just graphed it using (x,y) coordinates and looked at the boundaries.
 
  • #4
SammyS
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it's the function of the y interval --> [tex]\sqrt{}2x-x^2[/tex]
I didn't convert it polar coordinates, I just graphed it using (x,y) coordinates and looked at the boundaries.
You need it in polar coordinates to get your integration limits for the iterated integral in polar coordinates. The upper limit of integration for r is a function of θ .
 
  • #5
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You need it in polar coordinates to get your integration limits for the iterated integral in polar coordinates. The upper limit of integration for r is a function of θ .

OOOOH
Ok Imma try that.
 
  • #6
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So, I used x and y trig identities and got this
y-upper bound: r(theta) = 2cos(theta)
y-lower bound: r(theta) = 0

But How do I get theta bounds?
 
Last edited:
  • #7
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So, I used x and y trig identities and got this
y-upper bound: r(theta) = 2cos(theta)
y-lower bound: r(theta) = 0

But How do I get theta bounds?

Wait.... I already have the theta bounds: it's 0 to pi. For theta, I just look at the graph.
 
  • #8
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Wait.... I already have the theta bounds: it's 0 to pi. For theta, I just look at the graph.

ooops never mind. I finally got it. So, the final region D in polar coordinates is

r-upper: 2cos(theta)
r-lower:0

theta-upper:pi/2
theta-lower:0

i just used r = 2cos(theta) to solve for theta bounds.
 
  • #9
SammyS
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Yes! Looks good.
 

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