Find the area cut out of the cylinder x^2+z^2=81 by the cylinder x^2+y^2=81.

Question:

Find the area cut out of the cylinder {eq}x^2+z^2=81 {/eq} by the cylinder {eq}x^2+y^2=81 {/eq}.

Application of Double Integral

This problem is based on the application of double integrals. We have been given two different equations of cylinders, first, we are going to find the limits of integration and the function to be integrated followed by defining the limits to calculate the desired area.

Answer and Explanation:


The equations of cylinders we have are,

{eq}\displaystyle x^2+y^2=81 {/eq}

and,

{eq}\displaystyle x^2+z^2=81 {/eq}

So,

{eq}\displaystyle z=\sqrt{81-x^2} {/eq}

{eq}\displaystyle f(x,y)=\sqrt{81-x^2} {/eq}

{eq}\displaystyle f_x=\frac{-x}{\sqrt{81-x^2}} {/eq}

and,

{eq}\displaystyle f_y=0 {/eq}

We are going to assume that {eq}z>0 {/eq}

Thus, the required area is,

{eq}\displaystyle A=2\int \int \sqrt{(f_x)^2+(f_y)^2+1}dA {/eq}

{eq}\displaystyle A=2\int \int \sqrt{\left ( \frac{-x}{\sqrt{81-x^2}} \right )^2+(0)^2+1}dA {/eq}

{eq}\displaystyle A=2\int \int \sqrt{\left ( \frac{x^2+81-x^2}{{81-x^2}} \right )}dA {/eq}

On simplification, we get,

{eq}\displaystyle A=2\int \int {\left ( \frac{9}{\sqrt{81-x^2}} \right )}dA {/eq}

{eq}\displaystyle A=18\int \int {\left ( \frac{1}{\sqrt{81-x^2}} \right )}dA {/eq}


Now,

{eq}\displaystyle x^2+y^2=81 {/eq}

{eq}\displaystyle y=\pm\sqrt{81-x^2} {/eq}

Putting, {eq}y=0 {/eq}, we get,

{eq}\displaystyle x=\pm 9 {/eq}

Thus, the area integral now becomes:

{eq}\displaystyle A=18\int_{-9}^{9} \int_{-\sqrt{81-x^2}}^{\sqrt{81-x^2}} {\left ( \frac{1}{\sqrt{81-x^2}} \right )}dydx {/eq}

{eq}\displaystyle A=18\int_{-9}^{9} {\left ( \frac{1}{\sqrt{81-x^2}} \right )}\left ( y \right )_{-\sqrt{81-x^2}}^{\sqrt{81-x^2}}dx {/eq}

{eq}\displaystyle A=18\int_{-9}^{9} {\left ( \frac{1}{\sqrt{81-x^2}} \right )}\left ( 2\sqrt{81-x^2} \right )dx {/eq}

{eq}\displaystyle A=36\int_{-9}^{9} dx {/eq}

{eq}\displaystyle A=36\left ( x \right )_{-9}^{9} {/eq}

{eq}\displaystyle A=36(9+9) {/eq}

Thus, the area cut out is:

{eq}\displaystyle \boxed{\displaystyle A=648} {/eq}


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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