## Calculus: Early Transcendentals 8th Edition

$V=\frac{4}{3} \pi \left[ h^{2}+2ah \right] ^{\frac{3}{2}}$
{Step 1 of 5}: First, find the points of intersection of the curves $x^{2}-y^{2}=a^{2}$ or $x=\sqrt[]{a^{2}-y^{2}}$ and $x=a+h$ (where a>0,h>0) $\sqrt[]{a^{2}-y^{2}}= \left( a+h \right)$ $a^{2}-y^{2}= \left( a+h \right) ^{2}$ $y^{2}= \left( a+h \right) ^{2}-a^{2}$ $y^{2}=a^{2}+h^{2}+2ah-a^{2}$ $y^{2}=h^{2}+2ah$ $y^{2}=h \left( h+2a \right)$ $y= \pm \sqrt[]{h \left( h+2a \right) }$ {Step 2 of 5}: As an example, take $a=1$ and $h=1$ and sketch the curve {Step 3 of 5}: Rotate this shaded region about the y-axis. Rotate this to produce a solid. If we cut a Horizontal slice from this solid, we get a washer. The outer radius of the washer is $a+h$ The inner radius of the washer is $\sqrt[]{a^{2}-y^{2}}$ The cross-sectional area of the washer $A \left( y \right) = \pi \left[ \left( a+h \right) ^{2}- \left( \sqrt[]{a^{2}-y^{2}} \right) ^{2} \right]$ $= \pi \left[ a^{2}+h^{2}+2ah-a^{2}-y^{2} \right]$ $= \pi \left[ \left( h^{2}+2ah \right) -y^{2} \right]$ {Step 4 of 5}: The volume of the solid $V= \int _{-\sqrt[]{k \left( k+2a \right) }}^{+\sqrt[]{k \left( k+2a \right) }}A \left( y \right) dy$ Let $h \left( h+2a \right) =h^{2}+2ah=k$ {Step 5 of 5}: Then $V= \pi \int _{-\sqrt[]{k}}^{\sqrt[]{k}} \left( k-y^{2} \right) dy$ $= \pi \left[ ky-\frac{y^{3}}{3} \right] _{-\sqrt[]{k}}^{\sqrt[]{k}}$ [By the fundamental theorem of calculus, part 2] $= \pi \left[ \left( k\sqrt[]{k}-\frac{ \left( \sqrt[]{k} \right) ^{3}}{3} \right) - \left( -k\sqrt[]{k}+\frac{ \left( \sqrt[]{k} \right) ^{3}}{3} \right) \right]$ $= \pi \left[ 2k\sqrt[]{k}-\frac{2}{3} \left( \sqrt[]{k} \right) ^{3} \right]$ $= \pi \left[ 2 \left( \sqrt[]{k} \right) ^{3}-\frac{2}{3} \left( \sqrt[]{k} \right) ^{3} \right]$ $= \pi \left[ \frac{4}{3} \left( \sqrt[]{k} \right) ^{3} \right]$ $= \pi \left[ \frac{4}{3}k^{\frac{3}{2}} \right]$ $V=\frac{4}{3} \pi \left[ h^{2}+2ah \right] ^{\frac{3}{2}}$