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.



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:

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

  • $\begingroup$ 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

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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.