Picture of my project and all the work:
I’m working on a final project for my calculus AB class where we have to find the volume of a object while revolving it around the $x$-axis and using at least two integrals.
I’m doing the sucker as shown in the picture. I believe all my formulas are right but $43279.104 $mm$^3$ just sounds too big for the volume.
I was wondering if anyone could verify that my formulas are correct, mainly the one for the sphere (the second one under the axis of revolution part)
The dimensions of the sphere are, $43$mm diameter, $21.5$mm radius, $105.5$mm from $0$ to the midpoint/radius of the sphere.
The dimensions of the stick are, $5$mm diameter, $2.5$mm radius, length of $84$mm
Total length of sucker is $127$mm
Would really appreciate it as I want to do good on my final.
The intuition for volume is that it's measuring the number of small cubes (millimetre cubes) that would fit inside the object. As the cube's volume is the product of three numbers, volumes tend to get large quite quickly.
Notice that $10\ \text{mm}=1\ \text{cm}$ so $1000\ (10^3)\ \text{mm}^3=1\ \text{cm}^3$.
So your answer is $43.279\dots\ \text{cm}^3$.
Does it still sound too big?
Or, since your value is not (explicitly) derived from formulae for the volumes of cylinders and spheres, use this to check:
Sphere: $V=\frac{4}{3}\pi\times 21.5^3\approx 41600 \ \text{mm}^3$
Cylinder: $V=\pi\times 2.5^2\times 84\approx 1600 \ \text{mm}^3$
Sucker: $82657\pi/6\ \text{mm}^3 = 43279.104\dots \text{mm}^3$