find the volume of a solid in the first octant bounded above the plane z=5-2y and below by the rectangle in the xy-plane {(x,y):0<= x<=3,0<=y<=2}
i am lost about setting up the integral can any one help me???
Volume of a solid would be $\displaystyle \int_a^b\int_c^d f(x,y) \,dy\,dx$
In this case, $\displaystyle f(x,y)=z=5-2y$.
You're given restrictions on the region R: $\displaystyle \left\{R: \ 0\leq x \leq3, \ 0\leq y \leq 2\right\}$. This defines the x and y limits of integration.
Do you think you can set up the volume integral now?
--Chris
Generalising:
When you have a volume bounded above by z = f(x,y) and below by z = g(x,y), then $\displaystyle V = \int \int_{R_{xy}} f(x,y) - g(x,y) \, dx \, dy$ where $\displaystyle R_{xy}$ is the region in the xy-plane that the volume projects onto.
Note: In this question, $\displaystyle f(x,y) = 5 - 2y$ and $\displaystyle g(x,y) = 0$ so the volume is given by $\displaystyle V = \int \int_{R_{xy}} 5 - 2y \, dx \, dy$, the result given by Chris L T521.