Any help greatly appreciated.
Find the volume of the solid under the surface z=(8+x)(8+y) and above the region in the xy-plane given by (x^2)+(y^2) less than or equal to 36.
Hint: Use polar coordinates.
Thanks in advance.
Mike
Any help greatly appreciated.
Find the volume of the solid under the surface z=(8+x)(8+y) and above the region in the xy-plane given by (x^2)+(y^2) less than or equal to 36.
Hint: Use polar coordinates.
Thanks in advance.
Mike
In cylindrical polars we have:Originally Posted by JaysFan31
$\displaystyle
I=\int_{r=0}^6 \int_{\theta=0}^{2\pi} \int_{z=0}^{(8+r \cos(\theta))(8+r\sin(\theta)} 1\ dz\ d\theta\ r.dr $$\displaystyle =\int_{r=0}^6 \int_{\theta=0}^{2\pi} \int_{z=0}^{(8+r \cos(\theta))(8+r\sin(\theta)} r\ dz\ d\theta\ dr
$
Where we have used the fact that the volume element in cylindrical polars
is $\displaystyle r\ d\theta\ dr \ dz$, and changed the order of integration to suit the problem.
RonL