A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
Sketch the regions enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. (Do this on paper. Your instructor may ask you to turn in this graph.)
y=sqrt(x)
y=(1/3)x
x=16
i did this problem a long time ago, im currently studying for a math final i have tomorrow. how would you set up the integral?
anonymous
 5 years ago
Sketch the regions enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. (Do this on paper. Your instructor may ask you to turn in this graph.) y=sqrt(x) y=(1/3)x x=16 i did this problem a long time ago, im currently studying for a math final i have tomorrow. how would you set up the integral?

This Question is Closed

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Split into two integrals \[\int_0^{9}\sqrt{x}\frac{x}{3}\, dx+\int_9^{16}\frac{x}{3}\sqrt{x}\, dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do you know which boundaries to use for which integral though?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Here : http://www.wolframalpha.com/input/?i=plot+y%3Dsqrt%28x%29%2C+y%3Dx%2F3+x%3D0..16

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0The 9 is obtained by solving \(\sqrt{x}=\frac{x}{3}\) To review about this material, watch my video here: http://www.youtube.com/watch?v=xdEesoEqpQ

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah i understand to find the boundaries, we set the two functions equal to eachother and solve. i got 9 and 0 to be the two boundaries, i just didnt know where the x=16 that was given would fit into the integral

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would take a second look at this. It is important to hand draw these, label and everything to get a good feel. sq rt x and 1/3 x create an area but that is a smoke screen for your instructor to take off points. The loop where the three of them meet, apparently 9 to 16 is the area you want.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 16 of course is given in your problem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so there would only be one integral, not two ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Once you find that area (there is only one area) but the question is hinting to you to integrate along y, If you integrate along x, you would need to do two integrations because of how sq rt x and 1/3 x meet.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohhh okay i gotcha. im going to try to integrate with respect to y and see how that goes. thank you so much :D

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Are you serious chaguanas? integrating with respect to y? I am curios how would you set up that integral.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, after looking at it, this can be done with respect to x and it is the second part of integral above. But the question is unusually wordy and led me to believe it was a trick question. Usually the question is simply stated find the area bounded by the curves. I hope Anna sees this before it is too late.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Well if you integrate along the \(y\) you will need 3 integrals!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ha, ha. God help Anna.
Ask your own question
Sign UpFind more explanations on OpenStudy
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
 Engagement 19 Mad Hatter
 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.