Changing the Order of Integration in this Triple Integral

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SUMMARY

The discussion focuses on evaluating the triple integral ##\iiint_D (x^2+y^2)\mathrm{d}V## over the region ##D## defined by the curves ##y=x^2##, ##z=4-y##, and ##z=0##. The user presented six different orders of integration and sought validation for their correctness. Feedback indicated that the third and fourth orders of integration were incorrect, prompting a request for clarification on the proper bounds. Suggestions included improving the formatting of the equations for clarity and utilizing tools like Mathematica for verification.

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ainster31
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Homework Statement



Evaluate ##\iiint_D (x^2+y^2)\mathrm{d}V##, where ##D## is the region bounded by the graphs of ##y=x^2##, ##z=4-y##, and ##z=0##.

Homework Equations





The Attempt at a Solution



So after over at least an hour of thinking, I might have all 6 orders of integration. Can someone check my answer please?

1:
$$x=-\sqrt{y}\quad to\quad x=\sqrt{y}\\y=0\quad to\quad y=4\\z=0\quad to\quad z=4-y$$

2:
$$x=-\sqrt{y}\quad to\quad x=\sqrt{y}\\y=0\quad to\quad y=4-z\\z=0\quad to\quad z=4$$

3:
$$y=0\quad to\quad y=4-z\\x=-\sqrt{-z+4}\quad to\quad x=\sqrt{-z+4}\\z=0\quad to\quad z=4$$

4:
$$y=0\quad to\quad y=4-z\\x=-2\quad to\quad x=2\\z=0\quad to\quad z=-x^2+4$$

5:
$$z=0\quad to\quad z=4-y\\x=-2\quad to\quad x=2\\y=x^2\quad to\quad y=4$$

6:
$$z=0\quad to\quad z=4-y\\x=-\sqrt{y}\quad to\quad x=\sqrt{y}\\y=0\quad to\quad y=4$$

I've been informed 3 and 4 are incorrect. Does anyone happen to know the correct bounds so I can know where I went wrong?
 
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ainster31 said:

Homework Statement



Evaluate ##\iiint_D (x^2+y^2)\mathrm{d}V##, where ##D## is the region bounded by the graphs of ##y=x^2##, ##z=4-y##, and ##z=0##.

Homework Equations





The Attempt at a Solution



So after over at least an hour of thinking, I might have all 6 orders of integration. Can someone check my answer please?

1:
$$x=-\sqrt{y}\quad to\quad x=\sqrt{y}\\y=0\quad to\quad y=4\\z=0\quad to\quad z=4-y$$

2:
$$x=-\sqrt{y}\quad to\quad x=\sqrt{y}\\y=0\quad to\quad y=4-z\\z=0\quad to\quad z=4$$

3:
$$y=0\quad to\quad y=4-z\\x=-\sqrt{-z+4}\quad to\quad x=\sqrt{-z+4}\\z=0\quad to\quad z=4$$

4:
$$y=0\quad to\quad y=4-z\\x=-2\quad to\quad x=2\\z=0\quad to\quad z=-x^2+4$$

5:
$$z=0\quad to\quad z=4-y\\x=-2\quad to\quad x=2\\y=x^2\quad to\quad y=4$$

6:
$$z=0\quad to\quad z=4-y\\x=-\sqrt{y}\quad to\quad x=\sqrt{y}\\y=0\quad to\quad y=4$$

I've been informed 3 and 4 are incorrect. Does anyone happen to know the correct bounds so I can know where I went wrong?

Got two suggestions:

(1) Your results are poorly-formatted. What's being integrated with respect to what? Not easy to see and when you're asking for help you want to be as clear, concise, and to the point as possible as a courtesy to those helping you. Don't want to format it nicely huh?

(2) Do you have access to Mathematica or other where you can just check them yourself?
 

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