MHB Integral on plane inside a cylinder

[mathmari]

Hey!

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)

 

[HallsofIvy]

Using "dxdy" as the differential you are projecting down to the xy-plane. That is the circle x^2+ y^2= 4 which you can cover by taking x from -2 to 2 and, for each x, y from -\sqrt{4- x^2} to -\sqrt{4- x^2}:
\int_{-2}^2\int_{-\sqrt{4- x^2}}^{\sqrt{4- x^2}} (x^2+ y^2)(x+ y+ 4) dydx

Or, in polar coordinates:
\int_0^{2\pi}\int_0^2 r^2(rcos(\theta)+ rsin(\theta)+ 4)r drd\theta= \left(\int_0^{2\pi} cos(\theta)d\theta\right)\left(\int_0^2 r^3 dr\right)+ \left(\int_0^{2\pi} sin(\theta)d\theta\right)\left(\int_0^2 r^3dr\right)+ 8\pi\int_0^2 r^3 dr= 8\pi\int_0^2 r^3 dr

 

[I like Serena]

Hey!

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?

Yep. (Nod)

So, we get the following:
$$\iint_{\Sigma}(x^2+y^2)zdA=\iint_{D}(x^2+y^2)(4+x+y)dxdy$$

What happened to $\| \Sigma_x \times \Sigma_y \|$? (Wondering)

How can we continue? Do we use here cylindrical coordinates?

That is advisable yes.
As HallsofIvy showed, it's fairly straight forward in cylindrical coordinates.
Working out the cartesian expression is probably doable, but it becomes messy pretty quick. (Nerd)

 

[mathmari]

Thank you so much! (Yes)