Cynthia Bell

2021-12-26

Given a curve having an equation of ${y}^{2}=4x$.
a) Compute the area bounded by the curve and the line $y=4$ and $x=0$.
b) How far is the centroid of the curve bounded by the line $y=4$ and $x=0$ from the x-axis.
c) What is the volume generated by this curve if it is revolved about the x-axis.

turtletalk75

Given: - ${y}^{2}=4x$
To Find: - Area bounded by the curve $y=4,x=0$

Area $={\int }_{x=a}^{x=b}{\int }_{y={f}_{1}\left(x\right)}^{y={f}_{2}\left(x\right)}\text{dy dx}$
${\int }_{0}^{4}{\int }_{2\sqrt[\left(\right)]{x}}^{4}\text{dy dx}={\int }_{0}^{4}\left(4-2\sqrt[\left(\right)]{x}\right)\text{dx}$
$={\left[4x-\frac{2{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_{\left\{0\right\}}^{4}={\left[4x-\frac{4{x}^{\frac{3}{2}}}{3}\right]}_{\left\{0\right\}}^{4}$
$=\left(4\left(4\right)-\frac{4}{3}{4}^{\frac{3}{2}}\right)-\left(0\right)$
$=16-\frac{32}{3}=\frac{16}{3}$

Deufemiak7

b) To Find: - How For centroid of centroid
X-axis
i.e. we need to find y coordinute of centroid
$y=\frac{1}{\text{Area}}{\int }_{x=a}^{x=b}\frac{1}{2}\left(\left({f\left(x\right)}^{2}\right)-\left({g\left(x\right)}^{2}\right)\text{dx}$
$=\frac{1}{\frac{16}{3}}{\int }_{0}^{4}\frac{1}{2}\left[{4}^{2}-{\left(2\sqrt[\left(\right)]{x}\right)}^{2}\right]\text{dx}$
$=\frac{3}{32}{\int }_{0}^{4}\left(16-4x\right)\text{dx}$
$=\frac{3×4}{32}{\int }_{0}^{4}\left(4-x\right)\text{dx}$
$=\frac{3}{8}\left[4x-\frac{{x}^{2}}{2}\right]=\frac{3}{8}\left(8\right)=3$
 Centroid of curve is 3 from x-axis

karton

c) To Find: - Value generated by come f revolved about x-axis:
Volume of Solid revolved about the x-axis is given by
$V=\pi {\int }_{x=a}^{x=b}\left[f\left(x{\right)}^{2}\right]-\left[g\left(x{\right)}^{2}\right]\left(\right)\text{dx}$
whose y=f(x) is upper come f y=g(x) lower curve.
$\begin{array}{}\therefore V=\pi {\int }_{0}^{4}\left[{4}^{2}-{\sqrt[2]{x}}^{2}\right]\text{dx}\\ =\pi {\int }_{0}^{4}\left(16-4x\right)\text{dx}\\ =4\pi {\int }_{0}^{4}\left(4-x\right)\text{dx}\\ =4\pi \left[4x-\frac{{x}^{2}}{2}{\right]}_{0}^{4}\\ =4\pi \left[16-\frac{16}{2}\right]=4\pi \left(8\right)=32\pi \end{array}$

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