Question

Find the value of\(\iint_{S}{f}(x,y,z)dS \).

Short answer

The value of \(\iint_{S}{f}(x,y,z)dS \)is\( - 80\pi \).

Step-by-step solution

Concept of surface integral 

The surface integral is a generalization of multiple integrals that allows for surface integration. The surface integral is sometimes referred to as the double integral. We can integrate across a surface in either the scalar or vector fields for any given surface. The function returns the scalar value in the scalar field and function returns the vector value in the vector field.

“The surface integral of \(f\) over the surface \(S\) as\(\iint_S f(x,y,z)dS = \mathop {\lim }\limits_{\max \Delta {u_i},\Delta {v_j} \to 0} \sum\limits_{i = 1}^m {\sum\limits_{j = 1}^n f } \left( {P_{ij}^*} \right)\Delta {S_{ij}}\)where,\(\Delta {S_{ij}} \approx \left| {{r_u} \times {r_v}} \right|\Delta {u_i}\Delta {v_j}\)

Find the value of

As given data;

\(\ S be the hemisphere {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=4 \).

\(f(x,y,z)=g\left(\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}\right) andg(2)=-5 \).

The total and surface area of a sphere \(\left( \iint_{S}{d}S \right) is 4\pi {{r}^{2}} \)

\begin{aligned}& \iint_{S}{f}(x,y,z)dS\cong \iint_{S}{g}\left( \sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}\right)dS \\& \iint_{S}{f}(x,y,z)dS\cong \iint_{S}{g}(2)dS \\\end{aligned}

Substitute \( - 5\) for \(g\) (2);

\begin{aligned}& \iint_{S}{f}(x,y,z)dS\cong \iint_{S}{(-5)}dS \\ & \iint_{S}{f}(x,y,z)dS\cong (-5)\iint_{S}{d}S \\\end{aligned}

Substitute \(4\pi {{r}^{2}} fo \iint_{S}{d}S \);

\(\iint_{S}{f}(x,y,z)dS\cong(-5)\left(4\pi {{r}^{2}}\right) \)

Substitute 2 for\(r\);

\begin{aligned}& \iint_{S}{f}(x,y,z)dS\cong (-5)\left[ 4\pi {{(2)}^{2}} \right] \\& \iint_{S}{f}(x,y,z)dS\cong (-5)[16\pi ] \\& \iint_{S}{f}(x,y,z)dS\cong -80\pi\\\end{aligned}

Thus, the value of \(\iint_{S}{f}(x,y,z)dS \) is-80\(\pi \)..