Write and sketch the integral of intersection of $$y=\sqrt{x}, \ y=\sqrt{8-x}, \ y=1$$
Then find the volume of the shape rotating around the $x$-axis.
I think there is something wrong while writing the integral. Could you help for this question if possible?
Here is what I meant in the comment.
Shell Method:
$$V=2\pi \int_0^1 y\cdot [(8-y^2)-y^2] \ dy$$
Disk Method:
$$ V=\pi \int_0^1 (\sqrt{x})^2 \ dx + \pi \int_1^7 (1)^2 \ dx + \pi \int_7^8 (\sqrt{8-x})^2 \ dx$$
or (disk method, using symmetry)
$$ V=2\cdot \left[ \pi \int_0^1 (\sqrt{x})^2 \ dx + \pi \int_1^4 (1)^2 \ dx\right]$$