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Setup Double integral

  • Thread starter mugzieee
  • Start date
75
0
i have to setup a doble integral to find the volume of the solid bounded by the graphs of the equation.

x^2+z^2=1, and y^2+z^2=1
z=sqrt(1-x^2)
z=sqrt(1-y^2)

then substituting in z=sqrt(1-y^2) into x^2+z^2=1, i got y=x.

so when i setup a double integral
for the dy i get integral of(sqrt(1-y^2)) and the limits are from 0-x
and for the limits of he dx i get 0-1.

the way i got the lim its of x was by solving the equation x^2+z^2=1 for x by making z=0, but im not sure if thats right.

and finally when i integrate my double integral, im getting a wrong answer.
 
584
0
I'll reply to this with what I can help you with since no one else has replied yet. I remember posting a very similar problem a while ago but I've forgotten all about it now. :rolleyes: The following is just what I can come up with right now so its probably best to wait for further replies. I would also suggest including the answer which you seem to have.

When doing these sorts of problems I think you need to view projections wrt a single plane - for example yhe yz plane or the xy plane. This is just arbitrary but I would 'project' everything onto the xz plane. So you can 'ignore' the equation x^2 + z^2 = 1 for now. You also have y^2 + z^2 = 1 where y is the 'vertical' part of the coordinate system since the projection is onto the xz plane. Solving for y yields y = +/-sqrt(1-z^2).

[tex]
V = \int\limits_{ - 1}^1 {\int\limits_{ - \sqrt {1 - x^2 } }^{\sqrt {1 - x^2 } } {\int\limits_{ - \sqrt {1 - z^2 } }^{\sqrt {1 - z^2 } } {dydzdx} } }
[/tex]

Note: I checked on mathworldhttp://mathworld.wolfram.com/SteinmetzSolid.html) [Broken] and the volume is given by the integral above. I checked it so that I could verify my answer. Although I have little idea as to how this why the limits are the way that they are. I mean, if I had to guess I would've gone with those limits but I just don't understand them. The limits seem like they would only work if the cylinders intersected in perfect circles but this doesn't appear to be the case.

mugizee - don't worry about the paragraph above this, those are just things that have been troubling me. Also, you see that I have used a triple integral but it really just reduces to a double integral if that's what you really need to use. This is because the integrand is equal to one and so after the first integration it's equivalent to starting off with a double integral. I used a triple integral because I prefer to use them for volumes.

So [tex]V = \int\limits_{ - 1}^1 {\int\limits_{ - \sqrt {1 - x^2 } }^{\sqrt {1 - x^2 } } {\int\limits_{ - \sqrt {1 - z^2 } }^{\sqrt {1 - z^2 } } {dydzdx} } }[/tex]

[tex]
= \int\limits_0^{2\pi } {\int\limits_{ - \sqrt {1 - \sin ^2 \theta } }^{\sqrt {1 - \sin ^2 \theta } } {\int\limits_{ - \sqrt {1 - \cos ^2 \theta } }^{\sqrt {1 - \cos ^2 \theta } } {rdydrd\theta } } }
[/tex]...where x = sin(theta) and z = cos(theta), assigned this way because the y axis is the 'vertical' axis wrt to the chosen projection

[tex]
= \int\limits_0^{2\pi } {\int\limits_{ - \cos \theta }^{\cos \theta } {\int\limits_{ - \sin \theta }^{\sin \theta } {rdydrd\theta } } }
[/tex]

[tex]
= 2\int\limits_0^{2\pi } {\int\limits_{ - \cos \theta }^{\cos \theta } r } \sin \theta drd\theta
[/tex]

[tex]
= 2\int\limits_0^{2\pi } {\cos ^2 \theta \sin \theta d\theta }
[/tex]

[tex]
= - \frac{2}{3}\left[ {\cos ^3 \theta } \right]_0^{2\pi }
[/tex]

Ok there's an error somewhere since the 'answer' comes out to be zero. It should be 16/3. Perhaps someone else could give you some help with this because I can't get the answer.
 
Last edited by a moderator:
75
0
thanx for tryn Benny, but unfortanately the answer you gave was incorrect. the way the book does it is like this:
inner integral= sqrt(1-x^2)dy,0,x
outer integals limits are 0-1, dx
and the entire integral multiplied by two.

the final answer is 2/3

the equation they gave for the xy plane was simply y=x.

what i dont uderstand is why they chose the sqrt(1-x^2) equation for the z-equation, why couldnt they choose the other z=equation, or woul it have been the same? And why did they multiply the entire integral by two?
 
584
0
Hmm that's strange because the answer I got from mathworld is 16/3, perhaps I misread one of their answers. You can check if you want to, the relevant page is the link that I provided in my first post.

At the moment I just can't see why the book's solution works/doesn't work. Hopefully someone else can give you some help with this one. Sorry... :redface:
 

OlderDan

Science Advisor
Homework Helper
3,021
1
mugzieee said:
thanx for tryn Benny, but unfortanately the answer you gave was incorrect. the way the book does it is like this:
inner integral= sqrt(1-x^2)dy,0,x
outer integals limits are 0-1, dx
and the entire integral multiplied by two.

the final answer is 2/3

the equation they gave for the xy plane was simply y=x.

what i dont uderstand is why they chose the sqrt(1-x^2) equation for the z-equation, why couldnt they choose the other z=equation, or woul it have been the same? And why did they multiply the entire integral by two?
Benny's answer is correct for the entire region bounded by two intersecting cylinders. Your answer is correct for the region bounded by the intersecting cylinders and the planes y = 0 and y = x. Your integral is confined to the region y <= x in the first quadrant, which is one eighth of the region contained by the intersecting cylinders. Within your region of integration, z is limited by the smaller of the two values sqrt(1-x^2) and sqrt(1-y^2), which would be sqrt(1-x^2) since y <= x. The integral is multiplied by two because the range of z is symmetric about z = 0; the height of a volume element dV = zdxdy is 2*sqrt(1-x^2).
 
Last edited:
75
0
thanks, i get it now. that was a tricky problem...
 

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