The area of the region bounded by the circle | vedclass

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(B) The equation of the circle is $x^2 + y^2 = 16$, which can be written as $x^2 + y^2 = 4^2$. This is a circle centered at the origin $(0, 0)$ with a

Vedclass · Accepted Answer

$4$

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(B) The equation of the circle is $x^2 + y^2 = 16$, which can be written as $x^2 + y^2 = 4^2$. This is a circle centered at the origin $(0, 0)$ with a radius $r = 4$.In the first quadrant, the equation of the circle is $y = \sqrt{16 - x^2}$.The region is bounded by $x = 0$ and $x = 4$ in the first quadrant.The area $A$ is given by the integral: $A = \int_{0}^{4} y \, dx = \int_{0}^{4} \sqrt{16 - x^2} \, dx$.Using the standard integral formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a}) + C$, we get:$A = [\frac{x}{2} \sqrt{16 - x^2} + \frac{16}{2} \sin^{-1}(\frac{x}{4})]_{0}^{4}$.$A = [\frac{4}{2} \sqrt{16 - 16} + 8 \sin^{-1}(1)] - [0 + 8 \sin^{-1}(0)]$.$A = [0 + 8(\frac{\pi}{2})] - [0 + 0] = 4\pi$.Thus, the area is $4\pi$ sq. units.

The area of the region bounded by the circle $x^2+y^2=16$ and the lines $x=0$ and $x=4$ in the first quadrant is . . . . . . sq. units. (in $\pi$)

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