# Thread: The use of double and surface integrals: two problems

1. ## The use of double and surface integrals: two problems

Hello, boys , help "blonde" solve two problems on the double, triple and surface integrals, please.

#1. Find the center's coordinates of the gravity of a homogeneous surface:

$\displaystyle Z = \sqrt {a^2 - x^2 - y^2 } {\text{ }}\left( {x \geqslant 0,{\text{ }}y \geqslant 0,{\text{ }}x + y \leqslant a} \right).$

#2. Calculate the surface integral of the first kind of surface $\displaystyle S$,

$\displaystyle \iint\limits_S {\left( {5x - 8y - z} \right)d\sigma ,}{\text{ }}P:{\text{ }}2x - 3y + z = 6.$

where $\displaystyle S$ - part of the plane $\displaystyle P$, which cut coordinates planes.

How should I start to solve these problems? Do I need to change the coordinates in the first problem?
Which a figure is a projection $\displaystyle S$ on the plane XOY?

2. Originally Posted by LaraSoft
#2. Calculate the surface integral of the first kind of surface $\displaystyle S$,

$\displaystyle \iint\limits_S {\left( {5x - 8y - z} \right)d\sigma ,}{\text{ }}P:{\text{ }}2x - 3y + z = 6.$

where $\displaystyle S$ - part of the plane $\displaystyle P$, which cut coordinates planes.

Which a figure is a projection $\displaystyle S$ on the plane XOY?
The projection of $\displaystyle S$ on the plane $\displaystyle XOY$ is a triangle $\displaystyle D$: $\displaystyle {\text{ }}x > 0,{\text{ }}y < 0,{\text{ }}2x - 3y > 6$ (the fourth quadrant).

We need use this formula to calculate your surface integral:

$\displaystyle \iint\limits_S {f\left( {x,y,z} \right)}d\sigma = \iint\limits_D {f\left( {x,y,z\left( {x,y} \right)} \right)\sqrt {1 + {{\left( {\frac{{dz}}{{dx}}} \right)}^2} + {{\left( {\frac{{dz}}{{dy}}} \right)}^2}} dxdy}.$

$\displaystyle P:{\text{ }}2x - 3y + z = 6{\text{ }} \Leftrightarrow z = 6 - 2x + 3y$

$\displaystyle {\left( {\frac{{dz}}{{dx}}} \right)^2} = {\left( {\frac{d}{{dx}}\left( {6 - 2x + 3y} \right)} \right)^2} = 4,{\text{ }}{\left( {\frac{{dz}}{{dy}}} \right)^2} = {\left( {\frac{d}{{dy}}\left( {6 - 2x + 3y} \right)} \right)^2} = 9$

So $\displaystyle \sqrt {1 + {{\left( {\frac{{dz}}{{dx}}} \right)}^2} + {{\left( {\frac{{dz}}{{dy}}} \right)}^2}} = \sqrt {1 + 4 + 9} = \sqrt {14}$

and $\displaystyle f\left( {x,y,z\left( {x,y} \right)} \right) = 5x - 8y - \left( {6 - 2x + 3y} \right) = 7x - 11y - 6$.

Finally, we have

$\displaystyle D:{\text{ }}x > 0,{\text{ }}y < 0,{\text{ }}2x - 3y > 6.$

$\displaystyle \iint\limits_S {\left( {5x - 8y - z} \right)d\sigma } = \sqrt {14} \iint\limits_D {\left( {7x - 11y - 6} \right)}dxdy =$

$\displaystyle = \sqrt {14} \int\limits_0^3 {dx} \int\limits_{\frac{2} {3}x - 2}^0 {\left( {7x - 11y - 6} \right)dy} = \frac{{10\sqrt {14} }} {9}\int\limits_0^3 {\left( {9 + 3x - 2{x^2}} \right)dx} = 25\sqrt {14} .$