Change of variable in triple integrals

[Smusko]

Homework Statement

Solve for the volume above the xy-plane and below the paraboloid z=1-x²/a²-y²/b²

I have gotten an answer that is close to the correct one, but I can't figure out where I am wrong.

Homework Equations

Solution: Volume is = ab$$\pi$$/2

The Attempt at a Solution

First I substituted
u=x/a,
v=y/b,
w=z,
That changed dV = ab*dudvdw = dV*

Now the domain looks like this: Above the uv-plane and below 1=u²+v²+w


I substitute again to Cylindrical coordinates.

u=r*cos($$\theta$$)
v=r*sin($$\theta$$)
w=t
The are element becomes dV* = abr*drd$$\theta$$dt

So now if I have TripleIntegral(abr)drdtd$$\theta$$
From that I get ab*(r²/2)*(t)*(theta)
Evaluate this over
0≤r≤1
0≤$$\theta$$theta≤2pi
0≤t≤1
and I get ab*pi but the solution section says it becomes ab*pi/2 and I can't figure out where I am wrong or made a mistake.

Sorry for the alternating use of greek symbols and text. Sometimes the Latex reference works for me, sometimes it don't.

 

[Coto]

There appears to be either an error with how you've written the problem, or your first substitution. Is the a and b in the original equation for the paraboloid supposed to be squared? You've proceeded as if this were the case.

 

[Smusko]

Thanks, you are right. a and b are supposed to be squared in the problem statement. I'll edit it right away.

 

[LCKurtz]

So now if I have TripleIntegral(abr)drdtd$$\theta$$
From that I get ab*(r²/2)*(t)*(theta)
Evaluate this over
0≤r≤1
0≤$$\theta$$theta≤2pi
0≤t≤1
and I get ab*pi but the solution section says it becomes ab*pi/2 and I can't figure out where I am wrong or made a mistake.

Sorry for the alternating use of greek symbols and text. Sometimes the Latex reference works for me, sometimes it don't.

But t doesn't go from 0 to 1. The upper limit depends on r and θ.

 

[Smusko]

You are right. I'l check into that. Thanks.

 

[Smusko]

Actually the height is dependent on t. It is a cylinder and R and theta only decides the area of the circle. So that can't be what's wrong.
What I'm not 100 percent sure of is the limit.

 

[LCKurtz]

Homework Statement

Solve for the volume above the xy-plane and below the paraboloid z=1-x²/a²-y²/b²

I have gotten an answer that is close to the correct one, but I can't figure out where I am wrong.

Homework Equations

Solution: Volume is = ab$$\pi$$/2

The Attempt at a Solution

First I substituted
u=x/a,
v=y/b,
w=z,
That changed dV = ab*dudvdw = dV*

Now the domain looks like this: Above the uv-plane and below 1=u²+v²+w


I substitute again to Cylindrical coordinates.

u=r*cos($$\theta$$)
v=r*sin($$\theta$$)
w=t
The are element becomes dV* = abr*drd$$\theta$$dt

So now if I have TripleIntegral(abr)drdtd$$\theta$$
From that I get ab*(r²/2)*(t)*(theta)
Evaluate this over
0≤r≤1
0≤$$\theta$$theta≤2pi
0≤t≤1
and I get ab*pi but the solution section says it becomes ab*pi/2 and I can't figure out where I am wrong or made a mistake.

Sorry for the alternating use of greek symbols and text. Sometimes the Latex reference works for me, sometimes it don't.

But t doesn't go from 0 to 1. The upper limit depends on r and θ.

Actually the height is dependent on t. It is a cylinder and R and theta only decides the area of the circle. So that can't be what's wrong.
What I'm not 100 percent sure of is the limit.

Look at what I have highlighted in red. Your t is the same as w. The upper limit isn't constant. It depends on the other two variables as given by the w in the red equation.

 

[Smusko]

Ahhhhhh. Now I see... I think. How do I do to form an expression for w then? Or get its limit.

 

[LCKurtz]

Ahhhhhh. Now I see... I think. How do I do to form an expression for w then? Or get its limit.

Well, you have renamed the variable z to w to t. You needn't have changed it in the first place, but never mind that. So whether you call it z or w or t, it goes from 0 to whatever you get when you solve the red equation for w or the original equation for z, expressed in terms of your final variables. As usual in triple integrals, z goes from the bottom surface to the top surface if you integrate it first.

 

[Smusko]

Well, you have renamed the variable z to w to t. You needn't have changed it in the first place, but never mind that. So whether you call it z or w or t, it goes from 0 to whatever you get when you solve the red equation for w or the original equation for z, expressed in terms of your final variables. As usual in triple integrals, z goes from the bottom surface to the top surface if you integrate it first.

Of course. How stupid of me.

I think I have solved it now. I tried again with the new limit and failed, but then I noticed that I by mistake took the square of something that should not be squared. When I correct that mistake it should work out.

The source of error in these calculations are huge.

Thank you LCKurtz and coto for your help. :)