What is the triple integral of z^2(x^2 + y^2) over a bounded cylindrical region?

[3soteric]

Homework Statement

Let W= {(x,y,z)| x^2 + y^2 ≤ 1, -1 ≤ z ≤ 1} (W is a bounded cylindrical region)

Evaluate the triple integral f(x,y,z)= z^2 x^2 + z^2 y^2 over W. Use cylindrical coordinates

Homework Equations

i don't see any relevant equations besides the obvious cylindrical coordinate equations.

The Attempt at a Solution

I have no clue what the problem is asking for to be honest? any help on how to get me started? thank you

 

[jedishrfu]

convert the function to cyl coordinates using x = r cos(theta) and y=r sin(theta) and z=z and what do you get?

Can you simplify it?

 

[3soteric]

convert the function to cyl coordinates using x = r cos(theta) and y=r sin(theta) and z=z and what do you get?

Can you simplify it?

yes sir! I am on it right now! thanks

 

[Dick]

Homework Statement

Let W= {(x,y,z)| x^2 + y^2 ≤ 1, -1 ≤ z ≤ 1} (W is a bounded cylindrical region)

Evaluate the triple integral f(x,y,z)= z^2 x^2 + z^2 y^2 over W. Use cylindrical coordinates

Homework Equations

i don't see any relevant equations besides the obvious cylindrical coordinate equations.

The Attempt at a Solution

I have no clue what the problem is asking for to be honest? any help on how to get me started? thank you

Express your function and limits in term of the cylindrical coordinates. Change (x,y,z) to (r,θ,z) and don't forget to change dxdydz to its cylindrical equivalent. Then integrate. That's what it is asking you.

 

[3soteric]

f(x,y,z)= z^2 (r^2 cos^2theta) + z^2 (r^2 sin^2 theta)

 

[3soteric]

thank you dick! I am on it right now ! ill post results

 

[3soteric]

can i integrate the first limit then convert to polar? or do i have to flat out use cylindrical coordinates?

 

[Dick]

can i integrate the first limit then convert to polar? or do i have to flat out use cylindrical coordinates?

There's not that much of a difference. "cylindrical" is just "polar" plus the extra coordinate z. Use what you know about polar.

 

[3soteric]

if i draw the xz and yz traces, will it be a horizontal line at 1 and at -1? then of course the xy trace is a circle with r=1

 

[3soteric]

i got the following limits, r ≤ 1,
0≤ theta ≤ 2pi ,

-1≤z≤1

thats in polar right? now do i integrate? rdrdzdtheta?

 

[Dick]

if i draw the xz and yz traces, will it be a horizontal line at 1 and at -1? then of course the xy trace is a circle with r=1

The intersection with the xz and yz planes are rectangles, as the problem said, it's a cylinder. I'm not sure why you worried about this.

 

[Dick]

i got the following limits, r ≤ 1,
0≤ theta ≤ 2pi ,

-1≤z≤1

thats in polar right? now do i integrate? rdrdzdtheta?

Those are the limits and that's the differential element alright. So what function will you integrate?

 

[3soteric]

Those are the limits and that's the differential element alright. So what function will you integrate?

the function i will integrate will be z^2 x^2 + x^2 y^2 which i need to convert so its
z^2 (r^2 cos^2theta) + z^2 (r^2 sin^2 theta)

i integrate that? if so do i integrate rdzdrdtheta or rdrdzdtheta? i can't find a limit for dz, is it just -1 to 1? i don't have to express in other variables ? if so then its

∫2pi-0 ∫1-0 ∫ 1-(-1) (the function) r dz dr dtheta?

 

[3soteric]

by the way thanks for helping dick

 

[Dick]

by the way thanks for helping dick

No problem. But you can simplify the function you are integrating. Use a trig identity. It looks like you've got the rest ok.

 

[3soteric]

can i use sin^2+cos^2=1 then basically do z^2 r^2 + z^2 r^2 which is 2z^2r^2?

 

[Dick]

can i use sin^2+cos^2=1 then basically do z^2 r^2 + z^2 r^2 which is 2z^2r^2?

You got the identity right. But the result came out wrong. Give that another thought.

 

[3soteric]

i conclude z^2 (r^2 cos^2 theta + r^2 sin^2 theta) so i can factor the r^2 as well and I am left with (sin ^2 + cos ^2) and that's one

so z^2 r^2?

 

[Dick]

i conclude z^2 (r^2 cos^2 theta + r^2 sin^2 theta) so i can factor the r^2 as well and I am left with (sin ^2 + cos ^2) and that's one

so z^2 r^2?

Much better. Now put it all together.

 

[3soteric]

will do ! ill post back with results thank you

 

[3soteric]

my answer ended up being 4pi/9!

 

[Dick]

my answer ended up being 4pi/9!

Did you remember to include the r in r dz dr dtheta?

 

[3soteric]

Did you remember to include the r in r dz dr dtheta?

yep basically i did this


∫2pi-0 ∫1-0 ∫ 1-(-1) z^2 r^3 dz dr dtheta

i multiplied z^2r^2 by r so i got r^3
(first digit means upper limit for integral)

 

[3soteric]

 

[3soteric]

correction answer is pi/6 right?

 

[Dick]

correction answer is pi/6 right?

Yes, pi/6.