We see the function and a sketch of the volume in the following figure.
We use the shell method
![{\displaystyle V=2\pi \int h(z)z\,dz}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/b77c77045381d4171dc0c78754c790bacd87fb85)
to find the volume. Therefore we need to determine what is the height
and what is the integral variable
.
On the next figure we see, that the height is the inverse of the function
![{\displaystyle y=f(x)=1-x^{2}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/48ece284fd88497d68a740fb0a3d4b0b084f40b1)
And for this height the integral variable
is the distance from
to the rotating axis, which is
So, we find z = y + 2 and for the height we calculate
![{\displaystyle f(x)=y=1-x^{2}\ }](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/fc1abe03f16b565516e9bfc1e7a938d43732817d)
![{\displaystyle x^{2}=1-y\ }](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/8cb1192e6406ad893e94b97ff702aa13c0c3a8a9)
![{\displaystyle f^{-1}(y)=x={\sqrt {1-y}}.\ }](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/3d4ca628a647a565422a6c074397e61e4ae1622d)
For convenience, we calculate half of the volume and drop the left half. Then we can take
![{\displaystyle \displaystyle h(z)={\sqrt {1-y}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/51bcb019664c4742374ed4c0df7de2396beed6ec)
for z = y + 2.
The boundaries of the integral must be the left and right edge of the interval for
. Now we need to calculate the integral
![{\displaystyle {\begin{aligned}{\frac {1}{2}}V&=2\pi \int h(z)z\,dz\\&=2\pi \int _{0}^{1}(y+2){\sqrt {1-y}}\,dy\\&=2\pi \overbrace {\int _{0}^{1}\underbrace {y} _{=u}\underbrace {\sqrt {1-y}} _{=dv}\,dy} ^{\text{use integration by parts}}+2\pi \int _{0}^{1}2{\sqrt {1-y}}\,dy\\&=2\pi \left[y(1-y)^{\frac {3}{2}}(-{\frac {2}{3}})\right]_{0}^{1}+2\pi {\frac {2}{3}}\int _{0}^{1}(1-y)^{\frac {3}{2}}\,dy+2\pi 2\left[(1-y)^{\frac {3}{2}}\right]_{0}^{1}\\&=0+2\pi {\frac {2}{3}}\left[(1-y)^{\frac {5}{2}}(-{\frac {2}{5}})\right]_{0}^{1}-0+2\pi {\frac {4}{3}}\\&=2\pi {\frac {4}{15}}+2\pi {\frac {4}{3}}\\&={\frac {16\pi }{5}}\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9ff7a60fcf618697b98a372cdef803e1bca267ea)
So, we get for the volume
![{\displaystyle V={\frac {32\pi }{5}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/bb0ed476b95f5ea8ee61751c2c69f7dd500f57b3)