Express the volume of a right-circular cylindar of radius R and height h as an appropriate multiple integral, and evaluate this integral
soooo....yeah, not a clue
Let the tip of the cone be at the origin. Then the lower surface is $\displaystyle z = \frac{h}{R} \sqrt{x^2 + y^2}$ and the upper surface is z = h.
Therefore $\displaystyle V = \int \int_{R_{xy}} \int_{z = \frac{h}{R} \sqrt{x^2 + y^2}}^{z = h} \, dz \, dy \, dx $ where the region in the xy-plane is $\displaystyle x^2 + y^2 \leq R^2$.
The calculations are left for you. You should know what answer to expect.
Take as origin the centre of the base of the cylinder. The the base is the region:
$\displaystyle x^2+y^2 \le R^2$
we will calculate the volume above the first quadrant, which is 1/4 of the total volume so:
$\displaystyle V_{1st\ quad}=\int_{z=0}^h \int_{x=0}^R \int_{y=0}^{\sqrt{R^2-x^2}}\ dy \ dx\ dz$
CB
I'm struggling to see where you have obtained these values in the integral, i started the question by implementing the divergence theorem of Gauss but i wasn't sure if that was the way to go about it, so then started trying to find an answer using double integrals of volume. So because i don't know the methd, i certainly don't know where to start with plugging in values, even where they've come from!
No.
I did it because I'm blind and misread the question and gave you the answer to a harder one ..... (On the bright side, you'll probably have to answer a question like that sooner or later - I just brought it up sooner).
And for all worried members - my eyesight has returned (it's a miracle!)