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Evaluate the triple integral of $x=y^2$ over the region bounded by $z=x$, $z=0$ and $x=1$ My order of integration was $dx\:dy\:dz$.

I want to calculate the volume of this surface. I solved it for $dz\:dy\:dx$ and it was:

$$V=\int_0^1\int_{-\sqrt{x}}^{\sqrt{x}}\int_{0}^{x}\:dz\:dy\:dx$$

And for $dz\:dx\:dy$ would be this:

$$V=\int_{-1}^{1}\int_{y^2}^{1}\int_{0}^{x}dz\:dx\:dy$$

I tried to solve it and the result is this:

$$V=\int_{0}^{1}\int_{-\sqrt{x}}^{\sqrt{x}}\int_{z}^{1}dx\:dy\:dz + \int_{0}^{1}\int_{-\frac{1}{2}}^{\frac{1}{2}}\int_{y^2}^{1}dx\:dy\:dz$$

But i think its wrong please advice me the best solution .

I wanted to post the shape of this surface in 3-dimensional region but I couldn't because I am new user.

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What are the ▒ supposed to be? To get a proper integral sign with limits, enclose \int_0^1 in dollar signs to get $\int_0^1$ –  Ross Millikan Aug 5 '12 at 21:18

2 Answers 2

Integrating in three dimensions will give you a volume, not the area of a surface. Your region is not well defined in the first line-it is a triangle in the $xz$ plane but there is no restriction in the $y$ direction. If you want the region to be bounded by $x=y^2$ then your integral is correct, $V=\int_0^1\int_{-\sqrt{x}}^{\sqrt{x}}\int_{0}^{x}\:dz\:dy\:dx=\int_0^1\int_{-\sqrt{x}}^{\sqrt{x}}x\:dy\:dx=\int_0^12x\sqrt{x}\:dx=\frac 45 x^{\frac 52}|_0^1=\frac 45$. This is the triple integral of $1$ over that volume.

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yes it would be 4/5 for dzdydx & dzdxdy but for the dxdydz with that formula that i wrote its not the same so its wrong i don't know how to fix it –  shahin Aug 6 '12 at 0:37
    
for $dx \:dy \:dz$, the lower $x$ limit has to be the maximum of $y^2$ and $z$, you don't add the two integrals together. In your last line, you can't have $x$ in the limits of the $y$ integral as you have already integrated over $x$. In the second integral on that line, I don't know where the $y$ limits came from-we don't have any $\frac 12$'s around. –  Ross Millikan Aug 6 '12 at 0:50
    
yes i think it would be $$\int_0^1\int_{-1}^{1 }\int_1^{y^2}dxdydz$$ –  shahin Aug 6 '12 at 1:02
    
but its still wrong –  shahin Aug 6 '12 at 1:28

(to follow up on where this was left in August 2012):

enter image description here

Here is a graph of the region in question, made just incomplete enough to allow the interior to be viewed. The surface $ \ x = y^2 \ $ is a parabolic cylinder extending indefinitely in the $ \ z-$ directions; the surface $ \ z = x \ $ is an oblique plane; and $ \ z = 0 \ $ is, of course, the $ \ xy-$ plane. The volume is then a sort of wedge with a tilted "roof" and a parabolic "wall".

The integrals for the integration orders $ \ dz \ dy \ dx \ $ , which is sort of the "natural" order many people would use, and $ \ dz \ dx \ dy \ $ are shown in the other posts. One should not be too quick to arbitrarily re-arrange the order of integration in multiple integrals for a variety of reasons, sometimes because of the geometric configuration of the integration region, sometimes because of the integrands one would be left to grapple with.

In this problem, the amount of symmetry of the region permits us to choose alternative orders of integration without producing any "crisis". For the order $ \ dx \ dy \ dz \ $ , we are able to make use of the fact that one of the boundary surfaces is $ \ z \ = \ x \ $ . Since we want to "herd" the integration toward working in the variable $ \ z \ $ , we can carry out the integration in $ \ x \ $ from $ \ 0 \ $ to $ \ z \ $ . For the integration in $ \ y \ $ , we can therefore also "replace" $ \ x \ $ by $ \ z \ $ to express the relevant portion of the parabolic surface as $ \ z \ = \ y^2 \ $ . The integration limits on $ \ z \ $ become $ \ 0 \ $ to $ \ 1 \ $ , as they were for $ \ x \ $ .

We can now write the "re-ordered" integration as

$$ \int_0^1 \int_{-\sqrt{z}}^{\sqrt{z}} \int_0^z \ \ dx \ dy \ dz \ \ \ \text{or} \ \ \ 2 \ \int_0^1 \int_0^{\sqrt{z}} \int_0^z \ \ dx \ dy \ dz \ \ , $$

by exploiting the symmetry of the region about the $ \ xz-$ plane. Since this new integral simply looks like a "relabeled" version of the first integral in the original post, using the order $ \ dz \ dy \ dx \ $ , this will plainly give the same result for the volume of $ \ \frac{4}{5} \ $ .

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