pleasehelp20
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- Jan 23, 2018
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If someone could explain how to do this problem (in detail please) or show me how, I would greatly appreciate it.
1-2. The function \(\displaystyle f(x)\, =\, \sqrt{\strut R^2\, -\, x^2\,}\) has domain \(\displaystyle \left[-R,\, R\right].\) Assume \(\displaystyle 0\, \leq\, a\, \leq\, R.\) We wish to evaluate the following integral:
. . . . .\(\displaystyle \displaystyle \int_0^a\, \sqrt{\strut R^2\, -\, x^2\,}\, dx\)
(This corresponds to the shaded area in the graphic.)
Tragically, we do not know how to find an antiderivative for \(\displaystyle f(x).\) We will learn this later in the course. Instead, evaluate the integral geometrically by splitting the shaded area into the area of a triangle (which is pretty easy) plus the area of a circular sector (which will entail an inverse trig function).
I tried to figure it out in some scribbles on the side.
1-2. The function \(\displaystyle f(x)\, =\, \sqrt{\strut R^2\, -\, x^2\,}\) has domain \(\displaystyle \left[-R,\, R\right].\) Assume \(\displaystyle 0\, \leq\, a\, \leq\, R.\) We wish to evaluate the following integral:
. . . . .\(\displaystyle \displaystyle \int_0^a\, \sqrt{\strut R^2\, -\, x^2\,}\, dx\)
(This corresponds to the shaded area in the graphic.)
Tragically, we do not know how to find an antiderivative for \(\displaystyle f(x).\) We will learn this later in the course. Instead, evaluate the integral geometrically by splitting the shaded area into the area of a triangle (which is pretty easy) plus the area of a circular sector (which will entail an inverse trig function).
I tried to figure it out in some scribbles on the side.
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