Question

Evaluate the given integral by changing to polar coordinates. 49 − x2 − y2 darwhere r = (x, y) | x2 y2 ≤ 49, x ≥ 0

Answer 1

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

What is integration?

It is defined as the mathematical calculation by which we can sum up all the smaller parts into a unit.

We have the integral:

$\int\limits \int\limits_R {\sqrt{49-x^2-y^2}} \, dA$

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

$\int\limits \int\limits_R {\sqrt{49-x^2-y^2}} \, dA = \int\limits^{\dfrac{\pi}{2}}_{\dfrac{-\pi}{2}} \int\limits^7_0 {\sqrt{49-r^2]} \, rdrd\theta$

After solving the double integration, we will get:

$\int\limits \int\limits_R {\sqrt{49-x^2-y^2}} \, dA = \dfrac{343}{3} \pi$

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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