Find the volume under the surface z = xy above the triangle with vertices (1, 1, 0),(4, 1, 0),(1, 2, 0).
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The volume V lying beneath the surface z=f(x,y) and above the region
a≤x≤b,g1(x)≤y≤g2(x)V=∬Df(x,y)dA=∫ba∫g2(x)g1(x)f(x,y)dydxIn the triangle with vertices (0,1,0), (1,1,0) and (0,2,0), x lies between
0 and 1The y component could lie between y=1 and the line joining the points (1,1) and (0,2). And it could have the equation y=−x+2.
D={(x,y):0≤x≤1,1≤y≤−x+2}V=∬D(xy)dA=∫10∫−x+21xydydx
a≤x≤b,g1(x)≤y≤g2(x)V=∬Df(x,y)dA=∫ba∫g2(x)g1(x)f(x,y)dydxIn the triangle with vertices (0,1,0), (1,1,0) and (0,2,0), x lies between
0 and 1The y component could lie between y=1 and the line joining the points (1,1) and (0,2). And it could have the equation y=−x+2.
D={(x,y):0≤x≤1,1≤y≤−x+2}V=∬D(xy)dA=∫10∫−x+21xydydx
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