Question

The region under the curve `y=sqrtx` bounded by `0<=x<=4` is rotated about a) the x axis and b) the y axis. How do you sketch the region and find the volumes of the two solids of revolution?

Answer 1

See below

Explanation:

To graph the region:

sketch the curve `y=sqrtx` (This is the upper half of the parabola given by `x=y^2`) graph{y=sqrtx [-0.536, 9.33, -1.55, 3.383]}

Restrict the domain to `0 <= x <= 4` and
use the `x` axis as a lower boundary.

Shade in the regione under the curve `y = sqrtx` but above the `x`-axis for just the part of the `x`-axis where `0<=x<=4`

Revolve about the `x` axis

Use disks.

`V = int_0^4 pi(sqrtx)^2 dx`

`= pi[x^2/2]_0^4`

`= 8pi`

Revolve about the `y` axis

Use shells.

`V = int_0^4 2pisqrtx dx`

`= 2pi[(2x^(3/2))/3]_0^4`

`= (32pi)/3`