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Sagot :
The surface integral of the given surface [tex]z=\frac{2}{3}(x^\frac{3}{2}+y^\frac{3}{2})[/tex] in the intervals 0 ≤ x ≤ 4, 0 ≤ y ≤ 1 is 7.36615.
How to evaluate the surface integral?
To evaluate the surface integral, the formula applied is
[tex]S=\int\limits^b_a\int\limits^d_c {f(x,y)} \ dy\ dx[/tex]
⇒ S = [tex]\int\limits^b_a \int\limits^d_c {\sqrt{1+(\frac{dz}{dx})^2+(\frac{dz}{dy})^2 } } \, dy \, dx[/tex]
Where the differentiation is partial differentiation and f(x, y) = z.
Calculation:
The given function is
f(x, y) = [tex]z=\frac{2}{3}(x^\frac{3}{2}+y^\frac{3}{2})[/tex]
Then, applying the partial differentiation w.r.t x and y respectively,
dz/dx = [tex]\frac{2}{3}(\frac{3}{2})x^{1/2}[/tex] = [tex]x^{1/2}[/tex]
dz/dy = [tex]\frac{2}{3}(\frac{3}{2})y^{1/2}[/tex] = [tex]y^{1/2}[/tex]
Then,
we have surface integral formula as
S = [tex]\int\limits^b_a \int\limits^d_c {\sqrt{1+(\frac{dz}{dx})^2+(\frac{dz}{dy})^2 } } \, dy \, dx[/tex]
On substituting,
⇒ S = [tex]\int\limits^4_0 {\int\limits^1_0 {\sqrt{1+(x^{1/2})^2+(y^{1/2})^2} \, dy } \, dx[/tex]
⇒ S = [tex]\int\limits^4_0 {\int\limits^1_0 {\sqrt{1+x+y} \, dy } \, dx[/tex]
On integrating w.r.t y
⇒ S = [tex]\int\limits^4_0 {\frac{2}{3}(1+x+y)^{3/2}} \,|_0^1 dx[/tex]
Applying limits,
⇒ S = [tex]\int\limits^4_0 {\frac{2}{3}[(1+x+1)^{3/2}-(1+x+0)^{3/2} } \, dx[/tex]
⇒ S = [tex]\frac{2}{3} \int\limits^4_0 {[(x+2)^{3/2}-(x+1)^{3/2}}] \, dx[/tex]
Again applying the integration w.r.t x
⇒ S = [tex]\frac{2}{3}[\frac{2}{5}(x+2)^{5/2}-\frac{2}{5}(x+1)^{5/2}]|_0^4[/tex]
Applying the limits and simplifying,
⇒ S = [tex]\frac{4}{15}[(4+2)^{5/2}-(4+1)^{5/2}-(0+2)^{5/2}+(0+1)^{5/2}][/tex]
⇒ S = [tex]\frac{4}{15}[1+36\sqrt{6}-25\sqrt{5}-4\sqrt{2}][/tex]
∴ S = 7.36615
Therefore, the surface integral of the given function is 7.36615.
Learn more about the surface integral here:
https://brainly.com/question/4718231
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