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Evaluate the triple integral \iiint_W xz \, dV , where W = \{ (x,y,z) : \sqrt{x^2+y^2} \leq z...

Question:

Evaluate the triple integral {eq}\iiint_W xz \, dV {/eq}, where {eq}W = \{ (x,y,z) : \sqrt{x^2+y^2} \leq z \leq 9 \} {/eq}

Multiple Integral:

If we generalize the usual integral in one variable to a function of multiple variables in the higher dimension then integral obtained is called multiple integral.

For example:

{eq}\int \int \int _{R}f(x,y,z)dxdydz {/eq}

which is an integral over a three-dimensional region called triple integral.

Triple integral gives the volume of the curve.

Answer and Explanation:

Consider the given triple integral {eq}\int \int \int _{W}xzdV {/eq} where {eq}W=\left \{ (x,y,z):\sqrt{x^2+y^2}\le z\le 9 \right \} {/eq}

To evaluate the triple integral we convert it into cylindrical coordinate by substituting

{eq}x=r\cos \theta\\ y=r\sin \theta\\ z=z {/eq} and {eq}dV=dxdydz=rdrd\theta dz {/eq}

Our region of integration becomes {eq}r\le z\le 9 {/eq}.

The limit of integration will be

{eq}z\rightarrow r~to ~9\\ \theta \rightarrow 0~to~2\pi\\ r\rightarrow 0~to~3 {/eq}

{eq}\begin{align*} I&=\int_{\theta =0}^{2\pi}\int_{r=0}^{3}\int_{z=r}^{9}(r\cos \theta)zrdrd\theta dz\\ &=(\sin \theta)_{0}^{2\pi}\int_{r=0}^{3}\int_{z=r}^{9}r^2zdrd\theta dz\\ &=(\sin 2\pi -\sin 0)\int_{r=0}^{3}\int_{z=r}^{9}r^2zdrd\theta dz\\ &=(0-0)\int_{r=0}^{3}\int_{z=r}^{9}r^2zdrd\theta dz\\ &=0 \end{align*} {/eq}

So, the value of the triple integral is 0.


Learn more about this topic:

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Double Integrals: Applications & Examples

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14
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