# Triple integrals again

joemama69

## Homework Statement

evaluate the integral y2z2dV over W, which is the region bounded by x = 1 - y2 - z2 adn the plane x = 0

## The Attempt at a Solution

since x = 0, that makes y2 + z2 = 1, unit circle in the yz plane right?

so would the answer be the area of the circle times y2z2

## The Attempt at a Solution

Homework Helper
evaluate the integral y2z2dV over W, which is the region bounded by x = 1 - y2 - z2 adn the plane x = 0

so would the answer be the area of the circle times y2z2

uhh? the answer is a number … how can ∫∫∫ y2z2dV have any y or z in it? Homework Helper
Draw the region. You will see that it is a paraboloid with a maximum at y=z=0 and x=1. The paraboloid is cut off at x=0 so the region looks like the top of paraboloid. Which is not just a circle.

Homework Helper

Those integration limits are not correct. For A) if your integration order is dzdydx there will be a z left in your answer. Every other order of integration has a similar problem so it can't be right.

For B) you don't integrate over x,y and z, but over r, theta and x. This means you've to find limits for r, theta and x. The polar coordinates in this case would be $x=x, y=r \cos \theta, z=r \sin \theta,y^2+z^2=r^2$. Try to find the correct boundaries for both A and B. Realize the region is a paraboloid with an extremum at 1 on the x axis and cut off by the yz-plane.

I will leave it for you to decide which one is easier.

Homework Helper

You might try first swapping x and z so that the integrand is $x^2y^2dV$ and the bounding surfaces are $$\displaystyle z= 1- x^2- y^2$$ and z= 0.

In cylindrical coordinates that would be $(r^2cos^(\theta)r^2sin^2(\theta) rdrd\theta dz= r^5 sin^2(\theta)cos^2(\theta)$ and the boundaries $$\displaystyle z= 10- r^2$$ and z= 0.

joemama69

Well I was originally thinking polar was easier but i think ive changed my mind.

BTW im not using x2y2 and z = 1 - x2 - y2, z = 0

I need help setting up my limits

z.. 0 to 1 - x2 - y2

x... -(1-y2)1/2 to 1-y2)1/2

y... -1 to 1

whats right and whats wrong

Homework Helper

So you are integrating over the region bounded by $z= 1- x^2- y^2$ and z= 0?

I would recommend converting to cylindrical coordinates: $x= r cos(\theta)$ and $y= r sin(\theta)$ so that $z= 1- r^2$. Don't forget that the "differential of volume" in cylindrical coordinates is $dV= r dr d\theta$.

If you are required to do this in rectangular coordinates (or just want to), then, yes, your limits of integration are correct. Projecting the parboloid onto the z= 0 plane, you get the circle $x^2+ y^2= 1$. y going from -1 to 1 will cover that and, for any given y, x will go from $-\sqrt{1- y^2}$ to $\sqrt{1-x^2}$. Finally, for any given x and y, z going from 0 to $1- x^2- y^2$ will cover the solid.

joemama69
Ok, i tried it in cylindrical but i got my limts wrong

z is from 0 to 1-r2

r is from 0 to $$\sqrt{1-z}$$

theta is from 0 to pi

Homework Helper
You switched z and x around right? You can't have a z in your limits, it will mean you end up with a variable. The radii of the circular slices vary from the base r=1 to the top r=0. Also to integrate over the full circular slice you need to adjust your $\theta$ limits.

joemama69
im integrating over dz,dr,dQ Q = theta

dz = 0 to 1-r2

dr = 0 to 1

dQ = 0 to 2pi

$$\int\int\int$$r4cos2Q sin2Q r dz dr dQ

Homework Helper
Yes that should give you the same answer as the Cartesian method.

joemama69
ok i got stuck integrating

in integrated z and pluged in the limits and now im integrating r

$$\int$$(r5 - r7)cos2Q sin2Q dr = 1/24 .........

1/24 $$\int$$ cos2Q sin2Q dQ =

1/24 $$\int$$ ( - /sin2Q) sin2Q dQ =

1/24 $$\int$$ sin2Q - sin4Q dQ =

Now im stuck... my calculater gave me pi/4, but how do i do it by hand

Homework Helper
$sin^2(Q)- sin^4(Q)= sin^2Q(1- sin^2(Q))= sin^2(Q)cos^2(Q)$
Use trig identities $sin^2(Q)= (1/2)(1- cos(Q))$, $cos^2(Q)= (1/2)(1+ cos(Q))$. You may need to use those twice.

joemama69
i havent those identities in a while

1/24 $$\int$$cos2Q sin2Q dQ =

1/24 $$\int$$(1/2 + 1/2 cosQ)(1/2 - 1/2 cosQ) dQ =

1/96 $$\int$$(1 + cos2Q) dQ =

1/96 $$\int$$(1 + 1/2 + 1/2 cosQ) dQ =

1/96 $$\int$$(3/2 + 1/2 cosQ) dQ =

1/96 (3Q/2 + 1/2 sinQ)] from 0 to 2pi = pi/32

my calc said it was pi/4 do u see any problems

Homework Helper
I would personally start out with $\cos^2 \theta \sin^2 \theta=(\cos \theta \sin \theta)^2=\frac{1}{4}\sin^2 (2\theta)$. This way you can skip a few steps.

You used Halls identities which are incorrect the correct ones are $\cos^2x=\frac{1}{2}(1+\cos 2x),\sin^2x=\frac{1}{2}(1-\cos 2x)$. Note the double angle.

joemama69
ok when i integrated with the last identity i got pi/96 did u get this too

Homework Helper
I got pi/96.