Question

Question: To determine the value of the integralcorrect to four decimal places.

Short answer

The value of the integralcorrect to four decimal places is \(\underline {3.4893} \).

Step-by-step solution

Concept of Surface Integral and Formula Used.

A surface integral is a surface integration generalisation of multiple integrals. It can be thought of as the line integral's double integral equivalent. It is possible to integrate a scalar field or a vector field over a surface.

Definition used:

The value of d S is given by \(dS = \sqrt {1 + {{\left( {\frac{{dz}}{{dx}}} \right)}^2} + {{\left( {\frac{{dz}}{{dy}}} \right)}^2}} dxdy\), that is \(dS = \sqrt {1 + {{\left( {\frac{{dz}}{{dx}}} \right)}^2} + {{\left( {\frac{{dz}}{{\partial y}}} \right)}^2}} dxdy\).

Calculate the Surface Integral.

It is given that \(S:z = 3 - 2{x^2} - {y^2}\) lies above the

\(xy\frac{{\partial z}}{{\partial x}} = - 4x,\frac{{\partial z}}{{\partial y}} = - 2y\)

Substitute\(xy\frac{{\partial z}}{{\partial x}} = - 4x,\frac{{\partial z}}{{\partial y}} = - 2y\)in\(dS = \sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} dxdy\)and obtain the value of dS as follows:

\(\begin{array}{l}dS = \sqrt {1 + {{\left( {\frac{{\partial z}}{{dx}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} dxdy\\ = \sqrt {1 + {{( - 4x)}^2} + {{( - 2y)}^2}} dA\\ = \sqrt {1 + 16{x^2} + 4{y^2}} dA\end{array}\)

Substitute \(dS = \sqrt {1 + 16{x^2} + 4{y^2}} dA\) in

Substitute \(z = 3 - 2{x^2} - {y^2}\) in the above equation as shown below:

It is given that \({\rm{S}}\) is the part of the paraboloid \(z = 3 - 2{x^2} - {y^2}\) that lies above the x yplane \({\rm{S}}.\)is part of the paraboloid for which \(z > 0\).

Hence,

\(\begin{array}{l}3 - 2{x^2} - {y^2} > 0\\2{x^2} + {y^2} < 0\end{array}\)

This is the region inside an ellipse.

The region\({\rm{D}}\)is\(\left\{ {(x,y) \in D: - \frac{{\sqrt 3 }}{{\sqrt 2 }} \le x \le \frac{{\sqrt 3 }}{{\sqrt 2 }}, - \sqrt {3 - 2{x^2}} \le y \le \sqrt {3 - 2{x^2}} } \right\}\).

Thus, the integral in equation (1) can be rewritten as follows:

\(\int_{\sqrt {\frac{3}{2}} }^{\sqrt {\frac{3}{2}} } {\int_{ - \sqrt {3 - 2{x^2}} }^{\sqrt {3 - 2{x^2}} } {{x^2}} } {y^2}{\left( {3 - 2{x^2} - {y^2}} \right)^2}\sqrt {1 + 16{x^2} + 4{y^2}} dydx\)

Thus, the value of the given integral can be obtained as shown below:

Using the double integral calculator, obtain the value of the above integral as shown below:

From Figure 1, it can be observed that the value of the given integral is \(\underline {3.4893} \)