Show that the surface x = \frac {1}{\sqrt{(y^2+z^2)}}, where 1 less than or equal to x <...

Question:

Show that the surface {eq}x = \frac {1}{\sqrt{(y^2+z^2)}}, {/eq} where 1 less than or equal to {eq}x < \infty, {/eq} can be filled but not painted.

Surface and Volume Integrals

When given the expression for the surface {eq}x=f(y,z) {/eq} , the area of the surface can be calculated using the integral:

{eq}\begin{align} A(S)=\iint_D\sqrt{f_y^2(y,z)+f_z^2(y,z)+1}dA \end{align} {/eq}

where {eq}(y,z)\in D {/eq} and {eq}D {/eq} is the domain of the function in the y-z plane.

Besides, the volume will be:

{eq}\begin{align} V=\iint_Df(y,z)dA \end{align} {/eq}

Answer and Explanation:

We have the surface defined as:

{eq}x = \frac {1}{\sqrt{(y^2+z^2)}} {/eq}

and we have also:

{eq}\begin{equation} 1\le x<\infty\Rightarrow\;0<y^2+z^2\le 1 \end{equation} {/eq}

Now, showing that the surface can be filled is the same that show that the integral converges in the given domain and showing that it cannot be painted is the same that showing that the surface integral diverges.

Transforming to polar coordinates for the y-z plane we have that {eq}y^2+z^2=r^2 {/eq} and the function becomes:

{eq}\begin{equation} f(r,\theta)=\frac{1}{r} \end{equation} {/eq}

where {eq}r\in (0,1] {/eq} and {eq}\theta\in[0,2\pi] {/eq}

Therefore, we have that the integral for the volume is:

{eq}\begin{align} V&=\iint_D f(r,\theta)r\;dr\;d\theta\\ &=\int_{0}^{2\pi}\int_{0}^{1}\frac{1}{r}r\,dr\,d\theta\\ &=\int_{0}^{2\pi}\int_{0}^{1}\,dr\,d\theta=2\pi\\ \end{align} {/eq}

The integral converges and the surface can be filled.

In the case of the surface we have:

{eq}\begin{align} A(S)&=\iint_D\sqrt{f_y^2(y,z)+f_z^2(y,z)+1}dA\\ &=\iint_D\sqrt{\frac{1}{\left(y^2+z^2\right)^2}+1}dA\\ &=\int_{0}^{2\pi}\int_{0}^{1}\sqrt{\frac{r^4+1}{r^4}}r\,dr\,d\theta\\ &=2\pi\int_{0}^{1}\sqrt{\frac{r^4+1}{r^4}}r\,dr\\ &=2\pi\int_{0}^{1}\frac{\sqrt{r^4+1}}{r}\,dr\\ &=2\pi\left[\frac{\sqrt{r^4+1}}{2}-\frac{1}{2} \tanh ^{-1}\left(\sqrt{r^4+1}\right)\right]\Big|_{0}^{1}=\infty \end{align} {/eq}

Then the integral does not converge and the surface cannot be painted.


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Double Integrals & Evaluation by Iterated Integrals

from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 4
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