Calculating volume of body enclosed by surfaces (double integral)
I have to calculate the volume of body enclosed by following surfaces
$x^2+y^2=9$; (cylinder)
$x+y+z=10$;
$z=0$
My first task is construct body's projection on $0_{xy}$ plane, which is circle with R=3
Then, I have to construct inequality system to describe $D$At this point, I can't figure out whether I am supposed to convert into polar coordinates, giving me $$ D=\left\{ \begin{array}{c} 0⩽r⩽3 \\ 0⩽θ⩽π \\ \end{array} \right. $$
Or
Use Cartesian coordinates. I have browsed similar questions, and I have seen cylinder volume being solved both via Cartesian and polar coordinates.
In short, my question is whether $$\int_π^3dθ\int_0^3rsin(θ)\cdot rdr$$integral for above equations correct?
EDIT Okay, so I am doing the following
$$ D=\left\{ \begin{array}{c} 0⩽r⩽3 \\ 0⩽θ⩽2π \\ \end{array} \right. $$
$$V=\int_{A}z(x,y)\>dxdy=\int_0^{2\pi}\int_0^3 (10-r\cos\theta-r\sin\theta)rdr d\theta=\int_0^{2\pi}\int_0^3(10r-r^2\cos\theta-r^2\sin\theta)dr d\theta=\int_0^2 (5r\cdot3^2-\frac{3^3}{3}-\frac{3^3}{3})df=45\theta-9sin\theta-9(-cos\theta)=90\pi+9$$
When integrating, I get $90+9π$. Did I misunderstand something?
$\endgroup$1 Answer
$\begingroup$In cylindrical coordinates, the volume integral is
$$V=\int_{A}z(x,y)\>dxdy=\int_0^{2\pi}\int_0^3 (10-r\cos\theta-r\sin\theta)rdr d\theta=90\pi$$
Edit:
$$V=\int_0^3dr\int_0^{2\pi}(10r-r^2\cos\theta-r^2\sin\theta) d\theta$$$$=\int_0^3 (20\pi r-r^2\sin\theta|_0^{2\pi}+r^2\cos\theta|_0^{2\pi})dr$$$$=\int_0^3 (20\pi r+0+0)dr=20\pi\int_0^3rdr=90\pi$$
$\endgroup$ 4
