Answer:
Explanation:
You can use calculus to derive the equation for the volume of a cone.
if we take an circle at the tip of the cone the radius is 0 but what we want to do is integrate along the x axis from 0 up to the height of the cone H.
So, the equation becomes
![\int\limits^H_0 {\pi r^(2) } \, dx](https://img.qammunity.org/2023/formulas/mathematics/middle-school/fm4pq02hxh47bibwtus1vys95g2yktibqv.png)
r is related to x by the formula for a line; r = m*x+b
m is the rise over the run which is given by the maximum height H divided by the maximum radius R. So m = H/R and the intercept b = 0
now we have a formula for r in terms of x and with a little variable manipulation we can get a formula for x in terms of r
r = (R/H)*x
If we replace r by this formula in our integral we get:
![\int\limits^H_0 {\pi ((R/H)*x)^(2) } \, dx](https://img.qammunity.org/2023/formulas/mathematics/middle-school/mh1tstxpsp05fs1z4k9jmeeojrbs2au45a.png)
since
, R, and H are constants we can move them outside the integral:
![\pi (R^(2) /H^(2) )*\int\limits^H_0 {x^(2) } \, dx](https://img.qammunity.org/2023/formulas/mathematics/middle-school/g9jkbvnm62kbqzcprtvtq0di4asvc4tq6t.png)
the integral is easy. It is
which is equal to
![H^(3) /3](https://img.qammunity.org/2023/formulas/mathematics/middle-school/vp5o1nl4irlach1szvgkxi5cnw8tm4ld2m.png)
substituting back into the original formula we get the equation for the volume of a cone.
V =
![(\pi R^(2)H)/3](https://img.qammunity.org/2023/formulas/mathematics/middle-school/qvrv5jzd6g4wz7u6ek4cg7iax5nmtnauug.png)
Use this formula with the values given in the question to solve for R.
183.17 =
![\pi](https://img.qammunity.org/2023/formulas/mathematics/high-school/s4a9atp2j2mg6r59j0huycjcajdeqo7hpu.png)
*7/3
rearranging we get
= R
R ≅ 5