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Find the volume of the solid obtained by revolving region…

Find the volume of the...
Find the volume of the solid obtained by revolving region y=3-x and y=2*x^0.5, y=0 (0<=x<=3) about the y-axis.
Find the volume of the solid obtained by revolving region between x=y^2+2 and y=x-4, y=0 about the y-axis.
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Answered in 35 minutes by:
10/13/2017
Ryan
Ryan, Engineer
Category: Pre-Calculus
Satisfied Customers: 9,183
Experience: B.S. in Civil Engineering
Verified

Hi,

Thank you for the new request. I'll have these posted for you shortly.

Thanks,

Ryan

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Here is a link to the solutions for these problems:

Integrals2

As always, please feel free to ask if you have any questions about these solutions.

Thanks,

Ryan

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Customer reply replied 8 months ago
I cannot open file. Can you please use PDF?

Here is the link for the pdf version of this set:

Integrals2pdf

Again, please let me know if you have any trouble with this file.

In the future, I'll try to remember that you need a PDF version. A reminder from you would be helpful though, since I deal with a large number of customers and often can't remember precisely who needs which particular file type, etc.

Thanks,

Ryan

Ryan
Ryan, Engineer
Category: Pre-Calculus
Satisfied Customers: 9,183
Experience: B.S. in Civil Engineering
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Ryan and 87 other Pre-Calculus Specialists are ready to help you
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Customer reply replied 8 months ago
I still cannot open the file.
Customer reply replied 8 months ago
For "region between x=y^2+2 and y=x-4, y=0 about the y-axis" why the limits of integration are 0 and 2? Do we need to solve an equation y^2+2=y+4 or just look at the graph?

In many situations you would need to set the two functions equal to each other and solve for the value of y (or x, depending on the direction of integration) where the two curves intersect.

The main purpose of the graph is to visualize the region that is being revolved, and determine if the cross-sections will be solid disks, or washers.

In this particular case, however, it was obvious from the graph that the two functions intersected at y = 2, so explicitly solving for y wasn't necessary. If you wanted to show that, you would write:

y^2 + 2 = y + 4

y^2 - y - 2 = 0

(y - 2)(y + 1) = 0

y = 2, and y = -1

Clearly the y = -1 solution is outside the bounded region so that solution is discarded.

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Customer reply replied 8 months ago
Thank you

You're welcome.

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