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Q56E

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Found in: Page 807

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

To find: The area of the surface that creates by the intersection of two cylinders $${y^2} + {z^2} = 1$$ and $${x^2} + {z^2} = 1$$.

The area of the surface that creates by the intersection of two cylinders $${y^2} + {z^2} = 1$$ and $${x^2} + {z^2} = 1$$ is $$\underline {16}$$.

See the step by step solution

Step 1: Formula used:

Write the expression to find the surface area of the plane with the vector equation $${\rm{r}}(x,z)$$.

Here,

$${{\rm{r}}_x}$$ is the derivative of vector equation $${\rm{r}}(x,z)$$ with respect to the parameter $$x$$ and

$${{\rm{r}}_z}$$ is the derivative of vector equation $${\rm{r}}(x,z)$$ with respect to the parameter $$z$$.

Write the expression to find $${{\bf{r}}_x}$$.

$${{\rm{r}}_x} = \frac{\partial }{{\partial x}}({\rm{r}}(x,z))(2)$$

Write the expression to find $${{\bf{r}}_z}$$.

$${{\rm{r}}_z} = \frac{\partial }{{\partial z}}({\rm{r}}(x,z)){\rm{ (3) }}$$

As the required area of the surface is created by the interaction of two cylinders, the total surface area is the four times of surface area of any one face.

Calculate the area for the face of the surface that intersects the $$y$$-axis and multiply with 4 to find the total area of the required surface.

Rewrite the equation of first cylinder $${y^2} + {z^2} = 1$$ as follows.

$$y = \sqrt {1 - {z^2}}$$

Consider the $$x$$ and $$z$$ as parameters and parameterize the first cylinder $${y^2} + {z^2} = 1$$ as follows.

$$x = x,y = \sqrt {1 - {z^2}} ,z = z$$

Step 2: Write the vector equation of the first cylinder from the parametric equations as follows. $${\rm{r}}(x,z) = x{\rm{i}} + \left( {\sqrt {1 - {z^2}} } \right){\rm{j}} + z{\rm{k}}$$

Calculation of $${{\rm{r}}_x}$$ :

Substitute $$x{\rm{i}} + \left( {\sqrt {1 - {z^2}} } \right){\rm{j}} + z{\rm{k}}$$ for $${\rm{r}}(x,z)$$ in equation $$(2)$$,

\begin{aligned}{{\rm{r}}_x} &= \frac{\partial }{{\partial x}}\left( {x{\rm{i}} + \left( {\sqrt {1 - {z^2}} } \right){\rm{j}} + z{\rm{k}}} \right)\\ &= \frac{\partial }{{\partial x}}(x){\rm{i}} + \frac{\partial }{{\partial x}}\left( {\sqrt {1 - {z^2}} } \right){\rm{j}} + \frac{\partial }{{\partial x}}(z){\rm{k}}\\ &= (1){\rm{i}} + (0){\rm{j}} + (0){\rm{k}}\\ = (1){\rm{i}} + (0){\rm{j}} + (0){\rm{k}}\end{aligned}

Calculation of $${r_z}$$ :

Substitute $$x{\rm{i}} + \left( {\sqrt {1 - {z^2}} } \right){\rm{j}} + z{\rm{k}}$$ for $${\rm{r}}(x,z)$$ in equation (3),

\begin{aligned}{{\rm{r}}_z} &= \frac{\partial }{{\partial z}}\left( {x{\rm{i}} + \left( {\sqrt {1 - {z^2}} } \right){\rm{j}} + z{\rm{k}}} \right)\\ &= \frac{\partial }{{\partial z}}(x){\rm{i}} + \frac{\partial }{{\partial z}}\left( {\sqrt {1 - {z^2}} } \right){\rm{j}} + \frac{\partial }{{\partial z}}(z){\rm{k}}\\ &= (0){\rm{i}} + \frac{1}{{2\sqrt {1 - {z^2}} }}(0 - 2z){\rm{j}} + (1){\rm{k}}\\ &= (0){\rm{i}} - \frac{z}{{\sqrt {1 - {z^2}} }}{\rm{j}} + (1){\rm{k}}\end{aligned}

Calculation of $${{\rm{r}}_x} \times {{\rm{r}}_z}$$ :

Substitute (1) $${\rm{i}} + (0){\rm{j}} + (0){\rm{k}}$$ for $${{\rm{r}}_x}$$ and $$(0){\rm{i}} - \frac{z}{{\sqrt {1 - {z^2}} }}{\rm{j}} + (1){\rm{k}}$$ for $${{\rm{r}}_z}$$ in the expression $${{\rm{r}}_x} \times {{\rm{r}}_z}$$, $${{\rm{r}}_x} \times {{\rm{r}}_z} = ((1){\rm{i}} + (0){\rm{j}} + (0){\rm{k}}) \times \left( {(0){\rm{i}} - \frac{z}{{\sqrt {1 - {z^2}} }}{\rm{j}} + (1){\rm{k}}} \right)$$

Rewrite and compute the expression as follows.

$${{\rm{r}}_x} \times {{\rm{r}}_z} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\1&0&0\\0&{ - \frac{z}{{\sqrt {1 - {z^2}} }}}&1\end{array}} \right|$$

$$= \left| {\begin{array}{*{20}{c}}0&0\\{ - \frac{z}{{\sqrt {1 - {z^2}} }}}&1\end{array}} \right|{\rm{i}} - \left| {\begin{array}{*{20}{l}}1&0\\0&1\end{array}} \right|{\rm{j}} + \left| {\begin{array}{*{20}{c}}1&0\\0&{ - \frac{z}{{\sqrt {1 - {z^2}} }}}\end{array}} \right|{\rm{k}}$$

$$= (0 - 0){\rm{i}} - (1 - 0){\rm{j}} + \left( { - \frac{z}{{\sqrt {1 - {z^2}} }} - 0} \right){\rm{k}}$$

$$= (0){\rm{i}} - {\rm{j}} - \frac{z}{{\sqrt {1 - {z^2}} }}{\rm{k}}$$

Substitute $$(0){\rm{i}} - {\rm{j}} - \frac{z}{{\sqrt {1 - {z^2}} }}{\rm{k}}$$ for $${{\rm{r}}_x} \times {{\rm{r}}_z}$$ in equation (1),

Find the limits of $$x$$ and $$z$$ & from the equation of second cylinder $${x^2} + {z^2} = 1$$.

Rewrite the equation $${x^2} + {z^2} = 1$$ as follow.

$$x = \pm \sqrt {1 - {z^2}}$$

Therefore, the limits of $$x$$ are written as follows.

$$- \sqrt {1 - {z^2}} \le x \le \sqrt {1 - {z^2}}$$

Consider the limits of $$z$$ as follows.

$$- 1 \le z \le 1$$

Step 3: Apply the limits and rewrite the expression in equation (4) as follows.

$$A(S) = \int_{ - 1}^1 {\int_{ - \sqrt {1 - {z^2}} }^{\sqrt {1 - {z^2}} } {\left( {\frac{1}{{\sqrt {1 - {z^2}} }}} \right)} } dxdz$$

As the surface area is symmetrical to the $$x$$-and $$y$$-axes, apply the symmetry and rewrite the expression as follows.

\begin{aligned}A(S) &= (2)(2)\int_0^1 {\int_0^{\sqrt {1 - {z^2}} } {\left( {\frac{1}{{\sqrt {1 - {z^2}} }}} \right)} } dxdz\\ &= 4\int_0^1 {\left( {\frac{1}{{\sqrt {1 - {z^2}} }}} \right)} dz\int_0^{\sqrt {1 - {z^2}} } {(1)} dx\\ &= 4(x)_0^{\sqrt {1 - {z^2}} }\int_0^1 {\left( {\frac{1}{{\sqrt {1 - {z^2}} }}} \right)} dz\\ &= 4\left( {\sqrt {1 - {z^2}} - 0} \right)\int_0^1 {\left( {\frac{1}{{\sqrt {1 - {z^2}} }}} \right)} dz\end{aligned}

Simplify the expression as follows.

\begin{aligned}A(S) &= 4\left( {\sqrt {1 - {z^2}} } \right)\int_0^1 {\left( {\frac{1}{{\sqrt {1 - {z^2}} }}} \right)} dz\\ &= 4\int_0^1 {(1)} dz\\ &= 4(z)_0^1\\ &= 4(1 - 0)\end{aligned}

As the total surface area is 4 times the surface area of face of the surface that intersects the $$y$$-axis, the required total surface area is determined as follows.

$$${A_{{\rm{Total }}}} = 4A(S)$$$

Substitute 4 for $$A(S)$$,

$$16$$

Thus, the area of the surface that creates by the intersection of two cylinders $${y^2} + {z^2} = 1$$ and $${x^2} + {z^2} = 1$$ is$$16$$