Answer to Question #184941 in Civil and Environmental Engineering for John Prats

Question #184941

Determine the area of the surface generated by revolving a given curve about a given axis.

a.) y=x³ on (0,1), about the x-axis.

b. ) y=x² on (0,√2), about the y-axis.


1
Expert's answer
2021-05-31T03:43:02-0400

a)

y=x3y=x^3

dydx=3x2\frac{dy}{dx}=3x^2

Surface area of revolutionSurface\ area\ of\ revolution

2π01x31+(3x2)2 dx2\pi\int_{0}^{1}{x^3\sqrt{1+\left(3x^2\right)^2}\ dx}

let u=1+9x4let\ u=1+9x^4

du=36x3 dxdu=36x^3\ dx

2πudu362\pi\int{\sqrt u\frac{du}{36}}

π18(23)(u)32\frac{\pi}{18}\left(\frac{2}{3}\right)\left(u\right)^\frac{3}{2}

π27(1+9x4)32     (0, 1)\frac{\pi}{27}\left(1+9x^4\right)^\frac{3}{2}\ \ \ \ \ \left(0,\ 1\right)

π27((10)321)\frac{\pi}{27}\left(\left(10\right)^\frac{3}{2}-1\right)

1.13412π1.13412\pi


b)

y=x2y=x^2

x= yx=\ \sqrt y

dxdy=12y\frac{dx}{dy}=\frac{1}{2\sqrt y}

Surface are of revolutionSurface\ are\ of\ revolution

2π02y1+(12y)2 dy2\pi\int_{0}^{\sqrt2}{\sqrt y\sqrt{1+\left(\frac{1}{2\sqrt y}\right)^2}\ dy}

2π02y+14 dy2\pi\int_{0}^{\sqrt2}{\sqrt{y+\frac{1}{4}}\ dy}

2π (23)(y+14)32 (0, 2)2\pi\ \left(\frac{2}{3}\right)\left(y+\frac{1}{4}\right)^\frac{3}{2}\ (0,\ \sqrt2)

4π 3( (2+14)32(14)32)\frac{4\pi\ }{3}\left({\ \left(\sqrt2+\frac{1}{4}\right)}^\frac{3}{2}-\left(\frac{1}{4}\right)^\frac{3}{2}\right)

2.6958\pi\


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment