# Maya measures the radius and the height of a cone with 1% and 2% errors, respectively. She use these data to calculate the volume of the cone. What can Maya say about her percentage error in her volume calculation of the cone?

V_"actual"=V_"measured" pm4.05%, pm .03%, pm.05%

#### Explanation:

The volume of a cone is:

$V = \frac{1}{3} \pi {r}^{2} h$

Let's say we have a cone with r=1, h=1. The volume is then:

$V = \frac{1}{3} \pi {\left(1\right)}^{2} \left(1\right) = \frac{\pi}{3}$

Let's now look at each error separately. An error in $r$:

${V}_{\text{w/ r error}} = \frac{1}{3} \pi {\left(1.01\right)}^{2} \left(1\right)$

(pi/3(1.01)^2)/(pi/3)=1.01^2=1.0201=>2.01% error

And an error in $h$ is linear and so 2% of the volume.

If the errors go the same way (either too big or too small), we have a slightly bigger than 4% error:

1.0201xx1.02=1.040502~=4.05% error

The error can go plus or minus, so the final result is:

V_"actual"=V_"measured" pm4.05%

We can go further and see that if the two errors go against each other (one is too big, the other too small), they will very nearly cancel each other out:

1.0201(0.98)~=.9997=>.03% error and

(1.02)(.9799)~=.9995=>.05% error

And so, we can say that one of these values is correct:

V_"actual"=V_"measured" pm4.05%, pm .03%, pm.05%#