- #1

mathmari

Gold Member

MHB

- 5,049

- 7

I want to calculate $\iint_{\Sigma}(x^2+y^2)zdA$ on the part of the plane with equation $z=4+x+y$ that is inside the cylinder with equation $x^2+y^2=4$.

We can define the surface $\Sigma : D\rightarrow \mathbb{R}^3$ with $\Sigma (x,y)=(x,y,4+x+y)$, where $D$ is the space that is defined by $x^2+y^2\leq 4$, i.e. $D=\{(x,y)\mid x^2+y^2\leq 4\}$, right?

So, we get the following:

$$\iint_{\Sigma}(x^2+y^2)zdA=\iint_{D}(x^2+y^2)(4+x+y)dxdy$$

How can we continue? Do we use here cylindrical coordinates? (Wondering)