I want to find an integral which gives the volume of some region. First, I will give my solution and then the solution given by the manual.
The question is as below:
Calculate the volume of the region lying inside the cylinder $x^2+4y^2=4$, above the $xy-plane$ and below the plane $z=2+x$.
Here is what I have done: Set $x=2r\cos t$, $y= r\sin t$ so $x \le 2, y\le 1\implies 0\leq r\le 1$. Also we have $0\le t \le 2 \pi$. The image in my mind is that we have an ellipse on $xy-$plane, the parametrization we found above gives the domain and in order to find the volume we have to take the following integral :
$\displaystyle\int\int(2+x)dV$ = $\displaystyle\int_{0}^{2 \pi}\int_{0}^{1}(2+2rcost)rdrdt = \displaystyle\int_{0}^{2 \pi}(r^2+\cfrac{2r^3}{3}\cos t)dt = 2\pi$.
Here is what exactly the manual says: "Let $E$ be the elliptic disc bounded by $x^2+4y^2=4$. Then $E$ has area $\pi(2)(1) = 2 \pi$ square units. The volume of the region of $3-space$ lying above $E$ and below $z=2+x$ is $\displaystyle\int\int_E(2+x)dA = 2\int\int_EdA = 4\pi$ square units since $\displaystyle \int\int_E xdA= 0$ by symmetry."
Finally, I want to ask:
1)Where is my mistake?
2)How the manual solves the question, I mean how it thinks to say "$\displaystyle \int\int_E xdA= 0$ by symmetry" ?
Thank you.
First, you don't integrate over the volume, but over the area of the ellipse, so it's $dA$ not $dV$. But that's not where the error is. How did you get that the area is $r dr dt$? That is valid only for $x=r\cos t$ and $y=r\sin t$. Since that's not the change of variable that you did, your area element is $2r dr dt$.
The integral over a symmetric interval for an odd function is zero. When you integrate $x$ around the origin, you get $x^2/2$, which evaluated at any $+\alpha$ and $-\alpha$, are the same. In fact you already calculated this. The $\cos t$ term vanishes after the integration.