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Evaluate the triple integrals over the specified rectangular solid region.

R(x+2y+3z)dV, whereR={(x,y,z)0x4,1y5,and2z7}

Short Answer

Expert verified

R(x+2y+3z)dV=1720

Step by step solution

01

Step 1. Given information.

We have been given the triple integral:

R(x+2y+3z)dV, whereR={(x,y,z)0x4,1y5,and2z7}

We have to evaluate this over the specified rectangular solid regions.

02

Step 2. Evaluate.

By Fubini's theorem of triple integral :

R(x+2y+3z)dv=041527(x+2y+3z)dzdydx=041527(x+2y+3z)dzdydx=0415x27dz+2y27dz+327zdzdydx=0415x(z)27+2y(z)27+3z2227dydx=0415x(72)+2y(72)+3722222dydx=04155x+10y+1352dydx

03

Step 3. Integrate with respect to y.

Integrate with respect to y

=04155x+10y+1352dydx=045x15dy+1015ydy+135215dydx=045x(y)15+10y2215+1352(y)15dx=045x(51)+10522122+1352(51)dx=04[20x+120+270]dx=04[20x+390]dx

Integrate with respect to x

=2004xdx+39004dx=20x2204+390[x]04=20422+390[4]=160+1560=1720

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Most popular questions from this chapter

Evaluate each of the double integral in the exercise 37-54 as iterated integrals

R(x-ey)dAWhereR={(x,y)|-3x2,-2y2}

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