Answer to Question #189769 in Calculus for Abdul

Question

Express as a triple iterated integral the volume of the solid Q in the first octant bounded by the coordinate planes and the graphs of y^2+z^2=4 and 2z+x=4.Be certain to sketch the region Q for full credit

Answer 1

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\[ \begin{array}{l}
\mathrm{Re}gion\ \mathrm{:}y^{\mathrm{2}}+z^{\mathrm{2}}=\mathrm{4}\ \ \ \ \ and\ \ \ \mathrm{2}z+x=\mathrm{4} \\ 
In\ \ \ the\ \ first\ \ \ oc\mathrm{tan}t,\ x,y,z>0, \\ 
\therefore \ \ \ 0\le y\le \sqrt{\mathrm{4}-z^{\mathrm{2}}}\ \ \ ,\ \ 0\le z\le \mathrm{2}\ \ ,\ \ 0\le x\le \mathrm{4}-\mathrm{2}z \\ 
Volume\ \ =\ \ \mathop{\int\!\!\!\!\int\!\!\!\!\int}{dv} \\ 
Volume\ \ =\ \ \int^{\mathrm{2}}_0{\int^{\mathrm{4}-\mathrm{2}z}_0{\int^{\sqrt{\mathrm{4}-z^{\mathrm{2}}}}_0{dydxdz}}} \\ 
 \\ 
Volume\ \ =\ \ \int^{\mathrm{2}}_0{\int^{\mathrm{4}-\mathrm{2}z}_0{\int^{\sqrt{\mathrm{4}-z^{\mathrm{2}}}}_0{dydxdz}}} \\ 
 \\ 
Volume\ \ =\ \ \int^{\mathrm{2}}_0{\int^{\mathrm{4}-\mathrm{2}z}_0{\left(\sqrt{\mathrm{4}-z^{\mathrm{2}}}\right)dxdz}} \\ 
 \\ 
Volume=\int^{\mathrm{2}}_0{\left(\mathrm{4}-\mathrm{2}z\right)\left(\sqrt{\mathrm{4}-z^{\mathrm{2}}}\right)dz} \\ 
 \\ 
Set\ \ \ \ z=\mathrm{2sin}y\ \ ,\ \ \ \ \ dz=\mathrm{cos}y\ \ \ dy\ \  \\ 
 \\ 
Volume=\int^{\frac{\pi }{\mathrm{2}}}_0{\left(\mathrm{4}-\mathrm{4s}\mathrm{in}y\right)\left(\mathrm{2cos}y\right)\left(\mathrm{cos}y\right)dy} \\ 
 \\ 
Volume=\mathrm{8}\int^{\frac{\pi }{\mathrm{2}}}_0{\left(\mathrm{1}-\mathrm{sin}y\right)\left(\mathrm{co}{\mathrm{s}}^{\mathrm{2}}y\right)dy} \\ 
 \\ 
Volume=\mathrm{8}\int^{\frac{\pi }{\mathrm{2}}}_0{\left(\mathrm{co}{\mathrm{s}}^{\mathrm{2}}y\right)dy\ \ -\ \ \ }\mathrm{8}\int^{\frac{\pi }{\mathrm{2}}}_0{\left(\mathrm{sin}y\right)\left(\mathrm{co}{\mathrm{s}}^{\mathrm{2}}y\right)dy} \\ 
 \\ 
Volume=\mathrm{4}\int^{\frac{\pi }{\mathrm{2}}}_0{\left(\mathrm{1}+\mathrm{cos}\left(\mathrm{2}y\right)\right)dy\ \ +\ \ \ }\mathrm{8}\int^{\frac{\pi }{\mathrm{2}}}_0{\left(\mathrm{co}{\mathrm{s}}^{\mathrm{2}}y\right)d\left(\mathrm{cos}y\right)} \\ 
 \\ 
Volume=\mathrm{4}{\left(y+\frac{\mathrm{sin}\left(\mathrm{2}y\right)}{\mathrm{2}}\right)}^{\frac{\pi }{\mathrm{2}}}_0{}{}\ \ +\ \ {\frac{\mathrm{8}\left(\mathrm{co}{\mathrm{s}}^{\mathrm{3}}y\right)}{\mathrm{3}}}^{\frac{\pi }{\mathrm{2}}}_0 \\ 
 \\ 
Volume=\ \mathrm{2}\pi -\ \ \frac{\mathrm{8}}{\mathrm{3}} \end{array}
\] 




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