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enter image description hereLet S $\subset$ $R^2$ and let S be formed by :

  • the region inside the circle $x^2$+$y^2$=9

  • below the line y=x

  • above the x axis

  • laying to the right of x=1

Evaluate $\int xydA$.

I know how the region looks like.

The only problem I am having is how to draw the S region on a r and θ coordinate system.

Any help would be appreciated it.

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EDIT:

The way to transform Cartesian to polar is to use a paramrtrization in terms of the radius of the circle and the angle. Meaning:

$$ y = r \sin(\theta) $$ $$ x = r \cos(\theta) $$

Which is such that:

$$ x^2 + y^2 = r^2 $$

Using the formulae for changing of variables, we have also to multiply the Jacobian for the polar coordinates, which happens to be $ J(x,y) = r $ (the proof is quite simple).

You are integrating the region in first quadrant, in the intersection $x=y$ and the circumference starting from $x=1$.

Since $$ x = r \cos(\theta)$$

For $ x=1$ we have:

$$ 1 = r \cos(\theta) \Leftrightarrow r = \frac{1}{\cos(\theta)} = \sec(\theta) $$

Then for your particular case the region of integration $A$ is such that:

$$ \sec(\theta) \leq r \leq 3 $$ $$ 0 \leq \theta \leq \frac{\pi}{4}$$

Because the angle of $y=x$ is $\pi/4$. Then just integrate it:

$$ \iint_{A} r^3 \cos(\theta)\sin(\theta) dr d\theta$$

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  • $\begingroup$ I have attached a picture of how the S region looks like $\endgroup$
    – Hidaw
    Oct 12 '18 at 2:18
  • $\begingroup$ Do you have to integrate it in polar coordinates? This is a vertically simple region, I think it's easier to integrate in Cartesian. $\endgroup$ Oct 12 '18 at 2:26
  • $\begingroup$ Thank you so very much!!!! $\endgroup$
    – Hidaw
    Oct 12 '18 at 2:37
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You know how to draw it because the region is the same. The problem is to find equations for the boundaries. We have $x=r\cos \theta$ so you have that one. The circle is $r=3.$ Can you do the others?

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