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I'm studying to my test of Calculus and I'm not sure about the result I got. This is how I've done it:

Discover the interval of z: $$ 0 \leq z \leq x^2+y^2 $$

Setup the integral: $$ \iint_D \, \int_0^{x^2+y^2} \, dz \,\, dA $$

It equals to: $$ \iint_D x^2+y^2 \, dA $$

I've transformed the region $D$ to: $$ u = x \therefore x = u \\ v = y - 1 \therefore y = v + 1 \\ J = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1 $$

And setup the integral to: $$ \iint_{u^2+v^2=1} u^2+(v+1)^2 \, du \, dv = \iint_{u^2+v^2=1} u^2+v^2+2v+1 \, du \, dv $$

Using polar coordinates: $$ \int_0^{2\pi} \int_0^1 \left( r^2 + 2 r \sin \theta + 1 \right) r \, dr \, d\theta = \int_0^{2\pi} \int_0^1 r^3 + 2 r^2 \sin \theta + r \, dr \, d\theta $$

Solving: $$ \int_0^{2\pi} \frac{3}{4} + \frac{2}{3} \sin \theta \, d\theta = \frac{3\pi}{2} $$

So, is it right? I'm not sure about the transformation I've done.

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Set $D$ is $\{ (x,y,z ) \colon \ xy \leq z \leq x^2+y^2 \ \& \ x^2+y^2=2y\}$ so $\iiint_D \mbox{d}(x,y,z) \iint_{x^2+(y-1)^2 = 1} \left( x^2+y^2 - xy \right) \mbox{d}(xy)$. You calculated the volume of the set $S:= \{ (x,y,z ) \colon \ 0 \leq z \leq x^2+y^2 \ \& \ x^2+y^2=2y\}$ which is different. The substitution that you made is correct, you can apply it after correcting your $D$. – Frank Tessla Mar 3 '13 at 14:30
The value of $z$ when it crosses the plane $xy$ is $0$, right? So, I sincerely don't think these two regions are different. And it is confirmed when I've done the way you said and it returned me the same result – Rodrigo Siqueira Mar 3 '13 at 14:49
Yep, you are right and sets have the same measure $\frac{3}{2}\pi$ since $z=0$ when it crosses the plane $xy$. So your result is correct. – Frank Tessla Mar 3 '13 at 15:11
up vote 1 down vote accepted

Apart from a possible typo in $u^{2}+v^{2}=1$ instead of $u^{2}+v^{2}\leq 1$ in the definition of the region of integration $$ D=\left\{ (x,y)\in \mathbb{R}^{2}:x^{2}+(y-1)^{2}\leq 1\right\} =\left\{ (u,v)\in\mathbb{R}^{2}:u^{2}+v^{2}\leq 1\right\}$$

of the first two integrals in $u$ and $v$, everything is correct, in particular the transformation of coordinates $u=x,v=y-1$ and its Jacobian $J$.

$$ \iint_{u^{2}+v^{2}\leq 1}u^{2}+(v+1)^{2}\, du\, dv=\ldots =\int_{0}^{2\pi }\int_{0}^{1}r^{3}+2r^{2}\sin \theta +r\, dr\, d\theta. $$

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Oh, very well seen. I haven't noticed that – Rodrigo Siqueira Mar 3 '13 at 18:07
@Rodrigo Siqueira Perhaps I've noticed this detail because $u^2+v^2=1$ is a curve and you need to include its interior too, i.e. the disk $u^2+v^2\le 1$. – Américo Tavares Mar 3 '13 at 21:08

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