1. (Chap 7.2) Find the volume of a solid whose base is bounded by the circle centered at (0, 0) with radius 2,with a semicircle crossed section taken perpendicular to the x-axis.

Answered: 1. (Chap 7.2) Find the volume of a…

Calculus: Early Transcendentals

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1. (Chap 7.2) Find the volume of a solid whose base is bounded by the circle centered at (0, 0) with radius 2,
with a semicircle crossed section taken perpendicular to the x-axis.

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1. Given that the base of the solid is bounded by a circle centered at $(0, 0)$ with radius 2.

The general equation of a circle with center $(h, k)$ and radius r is given by ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ .

Hence, the equation of circle with center $(0, 0)$ and radius 2 is $x^{2} + y^{2} = 4$ .

The cross section taken perpendicular to the x-axis is a semicircle.

This cross section will look like a semicircle placed on the given base circle with only the diameter of the semicircle touching the base circle.

Hence, the diameter of the semicircle is given by the equation of the base circle $x^{2} + y^{2} = 4$ .

Note that, $x^{2} + y^{2} = 4 \Rightarrow y = \pm \sqrt{4 - x^{2}}$ .

The distance between the end points of the semicircle will give its diameter.

Both the end points will have the same x-coordinate as the cross section is taken perpendicular to the x-axis.

Hence, the distance between these end points is given by the difference between the y-coordinates of the end points.

One end point of the semicircle will have y-coordinate $y = \sqrt{4 - x^{2}}$ and the other end point will have y-coordinate $y = - \sqrt{4 - x^{2}}$ .

So, the diameter of the semicircle is $\sqrt{4 - x^{2}} - (- \sqrt{4 - x^{2}}) = 2 \sqrt{4 - x^{2}}$ .

Therefore, the radius of the semicircular cross section is $\sqrt{4 - x^{2}}$ .

The area of a semicircle of radius r is given by $\frac{{πr}^{2}}{2}$ .

Thus, the area of the required semicircular cross section is $\frac{π {(\sqrt{4 - x^{2}})}^{2}}{2} = \frac{π (4 - x^{2})}{2}$ .

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