# Thread: volume of hemisphere using Rieman's sums

1. ## volume of hemisphere using Rieman's sums

We have started getting volumes and areas using Reiman's sums and then turning them into definite integrals. I get some of this but there is one part when we take the antiderv that is not getting through to me!!!

If we have a hemsishere that has a radius of 7 cm^3 then the volume of a slice would be pi r^2 dh of slice

so our integral would be from 0 to 7 of pi(7^2 -h^2) dh

when this has its antiderv taken the book says it is

pi(7^2h - 1/3h^3 ) from 0 to 7 and then they say this is equal to

2/3 pi7^3

Please tell me how they get this since I thought the h was so small we do not use it and the other term would just equal to pi 7^2 (7)

In other words how do they get the 2/3?????????

2. Are you using washers?. We can generate the volume by revolving the upper half circle about the x-axis.

$\displaystyle f(x)=y=\sqrt{r^{2}-x^{2}}$

$\displaystyle V={\pi}\int_{-r}^{r}(r^{2}-x^{2})dx$

$\displaystyle {\pi}\left[r^{2}x-\frac{x^{3}}{3}\right]=\frac{4}{3}{\pi}r^{3}$

For a hemisphere, divide by 2. Just sub in your 7 to find the volume when r=7

The reason they got the 2/3 is because you want half of the sphere and 1/2 of 4/3 is 2/3.