integrate $int_0^1 int_1+y^2y int_z^y+z z, dx,dz,dy$

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My solution:



Integrate w.r.t $x$ and get $zx$, after inserting the boundaries of $x$ I get
$zy$.



Integrate w.r.t $z$ and get $y fracz^22$, after inserting the boundaries of $z$ I get $frac3y^3-2y^2-y2$.



Integrate w.r.t $y$ and get $frac3y^48 - fracy^33 - fracy^24$, after inserting the boundaries of $y$ I get $frac-524$.



Is my solution correct?




integration multivariable-calculus




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  • Looks correct to me. The following Mathematica command will check it for you: Integrate[Integrate[Integrate[z,x,z,y+z],z,1+y,2y],y,0,1].
    – Adrian Keister
    Jul 20 at 21:27














up vote
1
down vote

favorite












My solution:



Integrate w.r.t $x$ and get $zx$, after inserting the boundaries of $x$ I get
$zy$.



Integrate w.r.t $z$ and get $y fracz^22$, after inserting the boundaries of $z$ I get $frac3y^3-2y^2-y2$.



Integrate w.r.t $y$ and get $frac3y^48 - fracy^33 - fracy^24$, after inserting the boundaries of $y$ I get $frac-524$.



Is my solution correct?




integration multivariable-calculus




share|cite|improve this question





















  • Looks correct to me. The following Mathematica command will check it for you: Integrate[Integrate[Integrate[z,x,z,y+z],z,1+y,2y],y,0,1].
    – Adrian Keister
    Jul 20 at 21:27












up vote
1
down vote

favorite









up vote
1
down vote

favorite











My solution:



Integrate w.r.t $x$ and get $zx$, after inserting the boundaries of $x$ I get
$zy$.



Integrate w.r.t $z$ and get $y fracz^22$, after inserting the boundaries of $z$ I get $frac3y^3-2y^2-y2$.



Integrate w.r.t $y$ and get $frac3y^48 - fracy^33 - fracy^24$, after inserting the boundaries of $y$ I get $frac-524$.



Is my solution correct?




integration multivariable-calculus




share|cite|improve this question













My solution:



Integrate w.r.t $x$ and get $zx$, after inserting the boundaries of $x$ I get
$zy$.



Integrate w.r.t $z$ and get $y fracz^22$, after inserting the boundaries of $z$ I get $frac3y^3-2y^2-y2$.



Integrate w.r.t $y$ and get $frac3y^48 - fracy^33 - fracy^24$, after inserting the boundaries of $y$ I get $frac-524$.



Is my solution correct?





integration multivariable-calculus





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 20 at 21:25









Adrian Keister

3,61721533




3,61721533









asked Jul 20 at 21:20









kronos

909




909











  • Looks correct to me. The following Mathematica command will check it for you: Integrate[Integrate[Integrate[z,x,z,y+z],z,1+y,2y],y,0,1].
    – Adrian Keister
    Jul 20 at 21:27
















  • Looks correct to me. The following Mathematica command will check it for you: Integrate[Integrate[Integrate[z,x,z,y+z],z,1+y,2y],y,0,1].
    – Adrian Keister
    Jul 20 at 21:27















Looks correct to me. The following Mathematica command will check it for you: Integrate[Integrate[Integrate[z,x,z,y+z],z,1+y,2y],y,0,1].
– Adrian Keister
Jul 20 at 21:27




Looks correct to me. The following Mathematica command will check it for you: Integrate[Integrate[Integrate[z,x,z,y+z],z,1+y,2y],y,0,1].
– Adrian Keister
Jul 20 at 21:27










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    1 Answer
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    1 Answer
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    active

    oldest

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    up vote
    2
    down vote



    accepted










    Yes, your answer is correct. You have explained every step very clearly. Good Job!






    share|cite|improve this answer

























      up vote
      2
      down vote



      accepted










      Yes, your answer is correct. You have explained every step very clearly. Good Job!






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        Yes, your answer is correct. You have explained every step very clearly. Good Job!






        share|cite|improve this answer













        Yes, your answer is correct. You have explained every step very clearly. Good Job!







        share|cite|improve this answer













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        share|cite|improve this answer











        answered Jul 20 at 21:36









        Mohammad Riazi-Kermani

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