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$V={\int}_{a}^{b}\pi {[f(x)]}^{2}dx$

$V=\pi {\int}_{a}^{b}[f(x){]}^{2}dx$

$V=\pi {\int}_{0}^{2}{\left[\sqrt{4-{x}^{2}}\right]}^{2}dx$

$V=\pi {\int}_{0}^{2}4-{x}^{2}dx$

1.The original curve,

Imagine taking a quarter-circle cutout, and as you spin it around the

Every point on the rim of the quarter-circle moves in a circular path around the

The result is a solid that looks like half a ball where every point on the surface is equidistant from the center, located at the origin (0,0).

$V={\int}_{a}^{b}\u200a\pi {\left[f(x)\right]}^{2}dx$

$V=\pi {\int}_{a}^{b}\u200a[f(x){]}^{2}dx$

$V=\pi {\int}_{0}^{1}\u200a({x}^{2}{)}^{2}dx$

$V=\pi {\int}_{0}^{1}\u200a{x}^{4}dx$

$$\begin{array}{rl}& V=\pi {\left[\frac{{x}^{5}}{5}\right]}_{0}^{1}\\ & \\ & \\ & V=\pi [\left(\frac{{1}^{5}}{5}\right)-\left(\frac{{0}^{5}}{5}\right)]\\ & \\ & \\ & V=\pi [\left(\frac{1}{5}\right)-\left(\frac{0}{5}\right)]\\ & \\ & \\ & V=\frac{\pi}{5}\end{array}$$

2. Revolution: When this region is revolved around the

Here is the 3D illustration of the solid formed by revolving the region bounded by

This visualization shows a paraboloid, highlighting its shape and the curve that generates the solid.

$V={\int}_{a}^{b}\u200a\pi {\left[f(x)\right]}^{2}dx$

$V=\pi {\int}_{a}^{b}\u200a[f(x){]}^{2}dx$

$V=\pi {\int}_{1}^{4}\u200a(\sqrt{x}{)}^{2}dx$

$V=\pi {\int}_{1}^{4}\u200axdx$

$$\begin{array}{rl}& V=\pi {\left[\frac{{x}^{2}}{2}\right]}_{1}^{4}\\ & \\ & \\ & V=\pi [\left(\frac{(4{)}^{2}}{2}\right)-\left(\frac{(1{)}^{2}}{2}\right)]\\ & \\ & \\ & V=\pi [\left(\frac{16}{2}\right)-\left(\frac{1}{2}\right)]\\ & \\ & \\ & V=\frac{15\pi}{2}\end{array}$$

Region: The region is bounded by the curve $y=\sqrt{x}$ , the

Revolution: When this region is revolved around the

Solid of Revolution: The solid formed is a 3D shape known as a truncated cone, but with a curved rather than a straight side, narrowing as it extends from

Axis of Revolution: The

Cross-section: Circular disks whose radii vary as the square root of

$$V={\int}_{a}^{b}\pi {[f(x)]}^{2}dx$$

$$V=\pi {\int}_{a}^{b}[f(x){]}^{2}dx$$

$V=\pi {\int}_{-2}^{2}\u200a(\sqrt{4-{x}^{2}}{)}^{2}dx$

$V=\pi {\int}_{-2}^{2}\u200a(4-{x}^{2})dx$

$$\begin{array}{c}V=2\pi {[4x-\frac{{x}^{3}}{3}]}_{-2}^{2}\\ \\ \\ V=\pi [(4(2)-\frac{(2{)}^{3}}{3})-(4(-2)-\frac{(-2{)}^{3}}{3})]\\ \\ \\ V=\pi [(8-\frac{8}{3})-(-8-\frac{-8}{3})]\\ \\ \\ V=\pi [(\frac{24}{3}-\frac{8}{3})-(-\frac{24}{3}+\frac{8}{3})]\\ \\ \\ V=\pi [\left(\frac{16}{3}\right)-(-\frac{16}{3})]\\ \\ \\ V=\pi [\left(\frac{16}{3}\right)+\left(\frac{16}{3}\right)]\\ \\ \\ V=\frac{32\pi}{3}\end{array}$$

The original curve, $y=\sqrt{4-{x}^{2}}$ describes a semicircle with a radius of 2 centered at the origin, covering the domain from

Imagine taking a half-circle cutout, and as you spin it around the

Every point on the rim of the semicircle moves in a circular path around the

The result is a ball where every point on the surface is equidistant from the center, located at the origin (0,0).

$$V={\int}_{a}^{b}\pi {[f(x)]}^{2}dx$$

$$V=\pi {\int}_{a}^{b}[f(x){]}^{2}dx$$

The

$\begin{array}{rl}4-{x}^{2}& =0\\ & \\ \\ -{x}^{2}& =-4\end{array}$

$\begin{array}{rlr}\frac{-{x}^{2}}{-1}& =\frac{-4}{-1}& \\ & \\ \\ {x}^{2}& =4\end{array}$

$\begin{array}{rl}\sqrt{{x}^{2}}& =\sqrt{4}\\ \\ x& =\pm 2\end{array}$

$V=\pi {\int}_{-2}^{2}\u200a(4-{x}^{2}{)}^{2}dx$

$V=\pi {\int}_{-2}^{2}\u200a(4-{x}^{2})(4-{x}^{2})dx$

$V=\pi {\int}_{-2}^{2}\u200a(16-8{x}^{2}+{x}^{4})dx$

$\begin{array}{c}V=2\pi {[16x-8\frac{{x}^{3}}{3}+\frac{{x}^{5}}{5}]}_{-2}^{2}\\ & \\ & \\ V=\pi [(16(2)-8\frac{(2{)}^{3}}{3}+\frac{(2{)}^{5}}{5})-(16(-2)-8\frac{(-2{)}^{3}}{3}+\frac{(-2{)}^{5}}{5})]\\ & \\ & \\ V=\pi [(32-8\left(\frac{8}{3}\right)+\frac{32}{5})-(-32-8\left(\frac{-8}{3}\right)+\left(\frac{-32}{5}\right))]\\ & \\ & \\ V=\pi [(32-\frac{64}{3}+\frac{32}{5})-(-32+\frac{64}{3}-\frac{32}{5})]\end{array}$

Region: The region is bounded by the downward-facing parabola y = 4 - x

Solid of Revolution: When the area under the curve from

Axis of Revolution: The

Cross-section: Circular disks perpendicular to the

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