Use a double integral to find the volume of the indicated solid

In summary, the conversation discusses using a double integral to find the volume of a solid and the process of determining the correct limits for the integral. The final solution is determined to be ##\frac{20}{3}##.
  • #1
iRaid
559
8

Homework Statement


Use a double integral to find the volume of the indicated solid.
attachment.php?attachmentid=69225&stc=1&d=1398809837.png



Homework Equations





The Attempt at a Solution


I can't find what I did wrong, it seems like a simple problem...
$$\int_0^2 \int_0^x (4-y^{2})dydx=\int_0^2 4x-\frac{x^{3}}{3}dx$$
$$=2x^2-\frac{x^{4}}{12}|_0^2=8-\frac{16}{12}=\frac{20}{3}$$
 

Attachments

  • hw.png
    hw.png
    7.7 KB · Views: 1,497
Physics news on Phys.org
  • #2
iRaid said:

Homework Statement


Use a double integral to find the volume of the indicated solid.
attachment.php?attachmentid=69225&stc=1&d=1398809837.png



Homework Equations





The Attempt at a Solution


I can't find what I did wrong, it seems like a simple problem...
$$\int_0^2 \int_0^x (4-y^{2})dydx=\int_0^2 4x-\frac{x^{3}}{3}dx$$
$$=2x^2-\frac{x^{4}}{12}|_0^2=8-\frac{16}{12}=\frac{20}{3}$$

One of your limits for ##y## is wrong.
 
  • #3
Zondrina said:
One of your limits for ##y## is wrong.

I'm not seeing it, sorry.
 
  • #4
If you graph the region in the x-y plane, it should look something like this:

http://gyazo.com/aedf21fdd2006d58eabea7d5b3324065

Suppose you hold ##x## fixed and allow ##y## to vary. Then clearly from the above graph ##0 ≤ x ≤ 2## and ##x ≤ y ≤ 2##.

Try letting ##y## be fixed and allowing ##x## to vary now. Do you get the same result?
 
  • Like
Likes 1 person
  • #5
Zondrina said:
If you graph the region in the x-y plane, it should look something like this:

http://gyazo.com/aedf21fdd2006d58eabea7d5b3324065

Suppose you hold ##x## fixed and allow ##y## to vary. Then clearly from the above graph ##0 ≤ x ≤ 2## and ##x ≤ y ≤ 2##.

Try letting ##y## be fixed and allowing ##x## to vary now. Do you get the same result?

Ah right, I feel dumb now.

Thank you.
 

Related to Use a double integral to find the volume of the indicated solid

What is a double integral?

A double integral is a mathematical tool used to find the volume of a three-dimensional shape. It involves integrating a function over a two-dimensional region.

When would you use a double integral to find volume?

A double integral is used to find the volume of a solid when the shape cannot be expressed as a single function. This is often the case with more complex shapes such as cylinders, cones, and spheres.

What is the process for using a double integral to find volume?

The first step is to set up the integral by determining the limits of integration for both the x and y variables. This is done by analyzing the shape and its boundaries. Then, the function to be integrated is multiplied by the differential of both variables (dx and dy) and integrated over the given limits. Finally, the resulting integral is solved to find the volume of the solid.

Can a double integral be used to find the volume of any shape?

No, a double integral can only be used to find the volume of shapes with a constant cross-sectional area. If the cross-sectional area varies, a triple integral would be needed to find the volume.

Are there any limitations to using a double integral to find volume?

Yes, the shape must have a well-defined boundary and the limits of integration must be able to be determined. Additionally, the shape must have a constant cross-sectional area, as mentioned before. If these conditions are not met, a double integral cannot be used to find the volume.

Similar threads

  • Calculus and Beyond Homework Help
Replies
27
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
327
  • Calculus and Beyond Homework Help
Replies
10
Views
558
  • Calculus and Beyond Homework Help
Replies
1
Views
537
  • Calculus and Beyond Homework Help
Replies
34
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
383
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
807
Back
Top