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Let $E$ be the solid bounded by the plane $y=0$, the cylinder $z=1-x^2$, and the plane $y=z$. Set up the triple integral as an iterated integral.

My bounds so far are $z$ from $y$ to $1-x^2$

$y$ from $0$ to $1$

$x$ from $-1$ to $1$

Am I on the right track so far? I'm most unsure about my bounds for $y$ I got bounds for $x$ by setting $z=1-x^2$ equal to $y=z$, and solved for $x$ intercepts.

Michael Hardy
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user
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1 Answers1

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Any of the following six will do it, if my somewhat hasty computations are right. Maybe the main thing you should get from this is that these are done iteratively: First you decide the range of values for one variable; then decide, for any fixed value of that variable, what the range of values is for the next one, then decide for any fixed values of those two, what the range of values is for the third. And "first" means it's the outermost integral and "third" means the innermost. $$ \int_{-1}^1 \left( \int_0^{1-x^2} \left( \int_y^{1-x^2} 1 \,dz \right) \, dy \right) \, dx $$ $$ \int_{-1}^1 \left( \int_0^{1-x^2} \left( \int_0^z 1 \,dy \right) \, dz \right) \, dx $$ $$ \int_0^1 \left( \int_{-\sqrt{1-z}}^{\sqrt{1-z}} \left( \int_0^z 1\,dy \right) \, dx \right) \, dz $$ $$ \int_0^1 \left( \int_0^z \left( \int_{-\sqrt{1-z}}^{\sqrt{1-z}} 1\,dx \right) \, dy \right) \, dz $$ $$ \int_0^1 \left( \int_y^1 \left( \int_{-\sqrt{1-z}}^{\sqrt{1-z}} 1\,dy \right) \,dz \right)\, dy $$ $$ \int_0^1 \left( \int_{-\sqrt{1-y}}^{\sqrt{1-y}} \left( \int_y^{1-x^2} 1 \, dz \right) \, dx \right) \, dy $$

Michael Hardy
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