# Thread: triple integral in given region

1. ## triple integral in given region

Evaluate the triple integral where is the solid bounded by the paraboloid and .
what are the limit of z,y and x to integrate?

I got
$\displaystyle -sqrt(1-y^2) \leq z \leq sqrt(1-y^2)$
$\displaystyle -1 \leq y \leq 1$
$\displaystyle 0 \leq x \leq 9$

help me

2. hi

I perfer to transform the rectangular coordinates to polar ( cylindrial )

but before changing $\displaystyle dxdydz$ into $\displaystyle rdrd{\theta}dz$

we first exchange $\displaystyle x$ and $\displaystyle z$, it would be easier for us to solve .

Therefore , the boundary becomes $\displaystyle z = 9r^2 , z = 9$

and when $\displaystyle z = 9$ , $\displaystyle 9 = 9 r^2 \implies r=1$

the integral becomes $\displaystyle \int\int\int_{E_1} z rdrd{\theta}dz$

$\displaystyle = \int_0^{2\pi} \int_0^1 \int_{9r^2}^{9} zr dzdrd{\theta}$

$\displaystyle = 2\pi \int_0^1 r \frac{1}{2} (9^2 - 9^2 r^4 ) ~dr$

$\displaystyle = 81\pi \int_0^1 (r - r^5)dr = 27 \pi$

3. Originally Posted by zpwnchen
what are the limit of z,y and x to integrate?

I got
$\displaystyle -sqrt(1-y^2) \leq z \leq sqrt(1-y^2)$
$\displaystyle -1 \leq y \leq 1$
$\displaystyle 0 \leq x \leq 9$

help me

It should be

$\displaystyle \int_{-1}^1 \int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} \int_{9y^2 + 9z^2}^{9} x dxdydz$