The value of the integral is 343π/3 by changing to polar coordinates. √(49 − x2 − y2) dA where r = (x, y) | x2 y2 ≤ 49, x ≥ 0
It is defined as the mathematical calculation by which we can sum up all the smaller parts into a unit.
We have the integral:
[tex]\int\limits \int\limits_R {\sqrt{49-x^2-y^2}} \, dA[/tex]
Where, r = (x, y) | x2 y2 ≤ 49, x ≥ 0
The polar coordinate will be:
x = rcosθ
y = rsinθ
Where x²+y²= r²
Put the value of x and y in the integral, and the limits will be:
r²≤49 or 0≤r≤7, -π/2≤θ≤π/2 ( since x ≥0)
dA = rdrdθ
[tex]\int\limits \int\limits_R {\sqrt{49-x^2-y^2}} \, dA = \int\limits^{\dfrac{\pi}{2}}_{\dfrac{-\pi}{2}} \int\limits^7_0 {\sqrt{49-r^2]} \, rdrd\theta[/tex]
After solving the double integration, we will get:
[tex]\int\limits \int\limits_R {\sqrt{49-x^2-y^2}} \, dA = \dfrac{343}{3} \pi[/tex]
Thus, the value of the integral is 343π/3 by changing to polar coordinates. √(49 − x2 − y2) dA where r = (x, y) | x2 y2 ≤ 49, x ≥ 0
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