Sketch regions too
Inner: ò04-x xy dy=\allowbreak [ 1/2]xy2| y=04-x=[ 1/2]x( -4+x) 2
Completion: ò04[ 1/2]x( -4+x) 2dx=[ 1/2]ò04( x3-8x2+16x) dx=\allowbreak [ 32/3]
ò01òy=2x2( x+y) dydx=ò02ò0[ 1/2]y( x+y) dxdy=[ 5/3].
ò01òy=x3Öx( 48xy) dy dx=\allowbreakò01òx=y2y[ 1/3]( 48xy) dx dy=\allowbreak 5
Reverse order of integration.
Find Volume of solid
ò0[ 7/3]ò0-[ 3/2]x+[ 7/2](7-3x-2y) dy dx=ò0[ 7/3]ò07-3x( [(7-3x-z)/2]) dz dx=ò0[ 7/2]ò07-2y( [(7-2y-z)/3]) dz dy=\allowbreak [ 343/36]
ò-Ö2Ö2ò-Ö{2-x2}Ö{2-x2}( Ö{3-x2-y2}-[(x2+y2)/2]) dy dx=\allowbreak 2Ö3p-[ 5/3]p
Direct Computations in polar coordinates
Inner: ò13 re-r2 dr=\allowbreak -[ 1/2]e-9+[ 1/2]e-1 Using u=-r2 and du=-2r dr
Completion: ò0p/2ò13 re-r2 dr dq = \allowbreak -[ 1/4]e-9p+[ 1/4]e-1p
\diintD1 dA=ò02pò01+sin( q) 1 r dr dq = \allowbreak [ 3/2]p
\diintD1 dA=ò0pò04cos( q)1 r dr dq = \allowbreak 4p
\diintD1 dA=2( ò0[(p)/6]ò02sin( q) 1 r dr dq+ò[(p)/6][(p)/2]ò011 r dr dq) = \allowbreak [ 2/3]p-[ 1/2]Ö3
Convert from Cartesian ( x,y) to polar coordinates before integrating
\iintDf( x,y) dA=ò0[(p)/4]ò01f( rcos( q) ,rsin( q) ) r dr dq
V=ò-22ò-Ö{4-x2}Ö{4-x2}(4-x2-y2) dy dx=ò02pò02(4-r2) r dr dq = \allowbreak 8p
V=\diintD( Ö{25-x2-y2}-0) dA=2ò02pò35Ö{25-r2} r dr dq = \allowbreak [ 256/3]p
[This is a project problem but a hint is to write the equation of the cylinder in polar coordinates.]