# Thread: Volume Integrals?!?!

1. ## Volume Integrals?!?!

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

2. Originally Posted by Ash_underpar
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 $z = \frac{h}{R} \sqrt{x^2 + y^2}$ and the upper surface is z = h.

Therefore $V = \int \int_{R_{xy}} \int_{z = \frac{h}{R} \sqrt{x^2 + y^2}}^{z = h} \, dz \, dy \, dx$ where the region in the xy-plane is $x^2 + y^2 \leq R^2$.

The calculations are left for you. You should know what answer to expect.

3. Originally Posted by mr fantastic
Let the tip of the cone be at the origin. Then the lower surface is $z = \frac{h}{R} \sqrt{x^2 + y^2}$ and the upper surface is z = h.

Therefore $V = \int \int_{R_{xy}} \int_{z = \frac{h}{R} \sqrt{x^2 + y^2}}^{z = h} \, dz \, dy \, dx$ where the region in the xy-plane is $x^2 + y^2 \leq R^2$.

The calculations are left for you. You should know what answer to expect.
have you taken the equation of a cone because this is 1/3 the overall volume of the cylindar, so the volume of the cylindar is 3x the final value of the integral you have shown? thx

4. Originally Posted by Ash_underpar
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
Take as origin the centre of the base of the cylinder. The the base is the region:

$x^2+y^2 \le R^2$

we will calculate the volume above the first quadrant, which is 1/4 of the total volume so:

$V_{1st\ quad}=\int_{z=0}^h \int_{x=0}^R \int_{y=0}^{\sqrt{R^2-x^2}}\ dy \ dx\ dz$

CB

5. 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!

6. Originally Posted by Ash_underpar
have you taken the equation of a cone because this is 1/3 the overall volume of the cylindar, so the volume of the cylindar is 3x the final value of the integral you have shown? thx
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!)

7. Originally Posted by Ash_underpar
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!
Draw the cylinder in the way described by CaptainB. That will help you see where the integral terminals have come from.