A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Find the exact area of the surface obtained by rotating the curve about the x-axis. y =( (x^3)/3)+(1/4x) 1/2 <= x <= 1 Need Help!

  • This Question is Open
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    A solution using the Mathematica computer program is attached.

    1 Attachment
  2. ZeHanz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @robtobey: You probably misread the question. Not the volume of the body is needed, but its surface area. The formula for the surface area is: \(\int_{0.5}^1 2\pi f(x)\sqrt{1+(f'(x))^2}dx \) This would be: \( \int_{0.5}^1 2\pi(\frac{x^3}{3}+\frac{x}{4})\sqrt{1+(x^2+\frac{1}{4})^2}dx\)

  3. ZeHanz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I'm not confident if I can do this... Looking it up in WolframAlpha confirms my fears:

    1 Attachment
  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Refer to the attachment from the Mathematica v9 program.

    1 Attachment
  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @ZeHanz the integral isn't too bad, as it turns out: \[2\pi\int_{1/2}^1 \left(\frac{x^3}{3}+\frac{x}{4}\right)\sqrt{1+\left(x^2+\frac{1}{4}\right)^2}\,dx\] Pull out a factor of \(\frac{x}{3}\), rewrite, then substitute \(t=x^2+\dfrac{1}{4}\) so that \(\dfrac{1}{2}\,dt=x\,dx\), giving \[\frac{2\pi}{3}\int_{1/2}^1 x\left(x^2+\frac{3}{4}\right)\sqrt{1+\left(x^2+\frac{1}{4}\right)^2}\,dx=\frac{\pi}{3}\int_{1/2}^{5/4}\left(t+\frac{1}{2}\right)\sqrt{1+t^2}\,dt\] Next, set \(t=\tan v\), so \(dt=\sec^2v\,dv\), then the integral becomes \[\frac{\pi}{3}\int_{\arctan(1/2)}^{\arctan(5/4)}\left(\tan v+\frac{1}{2}\right)\sqrt{1+\tan^2v}\,\sec^2v\,dv\] Simplifying, \[\frac{\pi}{3}\int_{\arctan(1/2)}^{\arctan(5/4)}\left(\tan v+\frac{1}{2}\right)\sec^3v\,dv\] Upon expanding, we can easily deal with the remaining integration. \[\begin{align*}\int \tan v\sec^3v\,dv&=\int \tan v\sec v\sec^2v\,dv\\[1ex]&=\int \sec^2v\,d(\sec v)\\[1ex]&=\frac{1}{3}\sec v+C\\[2ex]\hline \int\sec^3v\,dv&=\sec v\tan v-\int \sec v\tan^2v\,dv\\[1ex] &=\sec v\tan v-\int \frac{\sin^2v}{\cos^3v}\,dv\\[1ex] &=\sec v\tan v-\int \frac{1-\cos^2v}{\cos^3v}\,dv\\[1ex] &=\sec v\tan v-\int (\sec^3v-\sec v)\,dv\\[1ex] 2\int\sec^3v\,dv&=\sec v\tan v+\int \sec v\,dv\\[1ex] \int\sec^3v\,dv&=\frac{\sec v\tan v+\ln|\sec v+\tan v|}{2}+C\end{align*}\]

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Typo, that first antiderivative should be \(\dfrac{1}{3}\sec^{\color{red}3}v+C\)

  7. ZeHanz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Well, that's impressive stuff, imo!

  8. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.