Question

Question: Express the integralas an iterated integral in six different ways, where \({\rm{E}}\) is the solid bounded by the given surfaces.

\({{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ = 9,x = - 2,x = 2}}\).

Short answer

All six integrals are the same.

\(\begin{array}{l}\int_{{\rm{ - 2}}}^{\rm{2}} {\int_{{\rm{ - 3}}}^{\rm{3}} {\int_{{\rm{ - }}\sqrt {{\rm{9 - }}{{\rm{z}}^{\rm{2}}}} }^{\sqrt {{\rm{9 - }}{{\rm{z}}^{\rm{2}}}} } {{\rm{f(x,y,z)dydzdx}}} } } \\\int_{{\rm{ - 2}}}^{\rm{2}} {\int_{{\rm{ - 3}}}^{\rm{3}} {\int_{{\rm{ - }}\sqrt {{\rm{9 - }}{{\rm{y}}^{\rm{2}}}} }^{\sqrt {{\rm{9 - }}{{\rm{y}}^{\rm{2}}}} } {{\rm{f(x,y,z)dzdydx}}} } } \\\int_{{\rm{ - 3}}}^{\rm{3}} {\int_{{\rm{ - 2}}}^{\rm{2}} {\int_{{\rm{ - }}\sqrt {{\rm{9 - }}{{\rm{z}}^{\rm{2}}}} }^{\sqrt {{\rm{9 - }}{{\rm{z}}^{\rm{2}}}} } {{\rm{f(x,y,z)dydxdz}}} } } \\\int_{{\rm{ - 3}}}^{\rm{3}} {\int_{{\rm{ - 2}}}^{\rm{2}} {\int_{{\rm{ - }}\sqrt {{\rm{9 - }}{{\rm{y}}^{\rm{2}}}} }^{\sqrt {{\rm{9 - }}{{\rm{y}}^{\rm{2}}}} } {{\rm{f(x,y,z)dzdxdy}}} } } \\\int_{{\rm{ - 3}}}^{\rm{3}} {\int_{{\rm{ - }}\sqrt {{\rm{9 - }}{{\rm{z}}^{\rm{2}}}} }^{\sqrt {{\rm{9 - }}{{\rm{z}}^{\rm{2}}}} } {\int_{{\rm{ - 2}}}^{\rm{2}} {{\rm{f(x,y,z)dxdydz}}} } } \\\int_{{\rm{ - 3}}}^{\rm{3}} {\int_{{\rm{ - }}\sqrt {{\rm{9 - }}{{\rm{y}}^{\rm{2}}}} }^{\sqrt {{\rm{9 - }}{{\rm{y}}^{\rm{2}}}} } {\int_{{\rm{ - 2}}}^{\rm{2}} {{\rm{f(x,y,z)dxdzdy}}} } } \end{array}\)

Step-by-step solution

Define integral

In mathematics, an integral assigns numbers to functions to express displacement, area, volume, and other concepts arising from linking infinitesimal data.

Explanation

Consider the following, \({{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ = 9}}\) and \({\rm{x = - 2,2}}\).

The good thing about this situation is that we'll always know where\({\rm{x}}\)is bounded,

\({\rm{ - 2}} \le {\rm{x}} \le {\rm{2}}\)

By solving for\({\rm{y}}\), we also know that\({\rm{y}}\)is bounded by,

\({\rm{ - }}\sqrt {{\rm{9 - }}{{\rm{z}}^{\rm{2}}}} \le {\rm{y}} \le \sqrt {{\rm{9 - }}{{\rm{z}}^{\rm{2}}}} \)

By solving for\({\rm{z}}\), we also know that\({\rm{z}}\)is bounded by,

\({\rm{ - }}\sqrt {{\rm{9 - }}{{\rm{y}}^{\rm{2}}}} \le {\rm{z}} \le \sqrt {{\rm{9 - }}{{\rm{y}}^{\rm{2}}}} \)

The six integrals are all the same.

\(\begin{array}{l}\int_{{\rm{ - 2}}}^{\rm{2}} {\int_{{\rm{ - 3}}}^{\rm{3}} {\int_{{\rm{ - }}\sqrt {{\rm{9 - }}{{\rm{z}}^{\rm{2}}}} }^{\sqrt {{\rm{9 - }}{{\rm{z}}^{\rm{2}}}} } {{\rm{f(x,y,z)dydzdx}}} } } \\\int_{{\rm{ - 2}}}^{\rm{2}} {\int_{{\rm{ - 3}}}^{\rm{3}} {\int_{{\rm{ - }}\sqrt {{\rm{9 - }}{{\rm{y}}^{\rm{2}}}} }^{\sqrt {{\rm{9 - }}{{\rm{y}}^{\rm{2}}}} } {{\rm{f(x,y,z)dzdydx}}} } } \\\int_{{\rm{ - 3}}}^{\rm{3}} {\int_{{\rm{ - 2}}}^{\rm{2}} {\int_{{\rm{ - }}\sqrt {{\rm{9 - }}{{\rm{z}}^{\rm{2}}}} }^{\sqrt {{\rm{9 - }}{{\rm{z}}^{\rm{2}}}} } {{\rm{f(x,y,z)dydxdz}}} } } \\\int_{{\rm{ - 3}}}^{\rm{3}} {\int_{{\rm{ - 2}}}^{\rm{2}} {\int_{{\rm{ - }}\sqrt {{\rm{9 - }}{{\rm{y}}^{\rm{2}}}} }^{\sqrt {{\rm{9 - }}{{\rm{y}}^{\rm{2}}}} } {{\rm{f(x,y,z)dzdxdy}}} } } \\\int_{{\rm{ - 3}}}^{\rm{3}} {\int_{{\rm{ - }}\sqrt {{\rm{9 - }}{{\rm{z}}^{\rm{2}}}} }^{\sqrt {{\rm{9 - }}{{\rm{z}}^{\rm{2}}}} } {\int_{{\rm{ - 2}}}^{\rm{2}} {{\rm{f(x,y,z)dxdydz}}} } } \\\int_{{\rm{ - 3}}}^{\rm{3}} {\int_{{\rm{ - }}\sqrt {{\rm{9 - }}{{\rm{y}}^{\rm{2}}}} }^{\sqrt {{\rm{9 - }}{{\rm{y}}^{\rm{2}}}} } {\int_{{\rm{ - 2}}}^{\rm{2}} {{\rm{f(x,y,z)dxdzdy}}} } } \end{array}\)

Therefore, all the six integrals are same.