Disc Method Finding Volume Trig Function

In summary, the region of the x-y plane between the line y=7, the curve y=3sin(x)+4, and for -pi/2<x<3pi/2 generates a solid that has a volume of pi*A(x)
  • #1
Chaoticoli
8
0

Homework Statement


Consider the region of the x-y plane between the line y=7, the curve y=3sin(x)+4 , and for
-pi/2<x<3pi/2

Find the volume of the solid generated by revolving this region about the line y=7.

I know that for this problem, I will be using the disk method (as the title states). I drew the graph and I find that there are two areas and since the graph is symmetric, the two areas are equal.

A picture of the graph is here: http://www.wolframalpha.com/input/?i=+the+curve+7=3*sin(x)+4+between+-pi/2<x<3pi/2

I know I must find the area of a slice. I am just not exactly sure how to find it. I know the equation for the area is A(x) = pi*r(x)^2

And then, you need to find the limits of integration and with those limits, the volume will be:

Volume = ∫ from a to b of pi*A(x)

Any help would be greatly appreciated ! :)
 
Physics news on Phys.org
  • #2
Obviously, I just need to find out what r(x) is and figure out if I should integrate with respect to x or y.
 
  • #3
Chaoticoli said:
since the graph is symmetric, the two areas are equal.
Not exactly. The area of interest is above the curve, so they're not so much equal as, in a sense, complementary.
Anyway, finding the area of regions in this graph won't help you.
I know I must find the area of a slice. I am just not exactly sure how to find it. I know the equation for the area is A(x) = pi*r(x)^2
Yes.
Can you visualise the rotation about y=7?
r(x) is the distance between two y values at x. What two y values?
 
  • #4
Yes. I visualize it being rotated, forming discs about the line y=7.

Is it the distance between y=3sin(x)+4 and y=7?

(3sin(x) + 4) - 7 = 3sin(x) -3 = r(x) ??
 
  • #5
Actually, I just figured it out :). I got 27pi^2 after integrating from -pi/2 to 3pi/2 of (3sin(x)-3)^2 dx
 

Related to Disc Method Finding Volume Trig Function

1. What is the disc method for finding volume using trigonometric functions?

The disc method is a mathematical technique used to find the volume of a solid of revolution, which is a three-dimensional shape formed by rotating a two-dimensional shape around an axis. In this case, the disc method specifically uses trigonometric functions to calculate the volume.

2. How does the disc method work?

The disc method works by breaking down the solid of revolution into infinitely thin discs, calculating the volume of each disc using the formula V = πr²h, and then summing up the volumes of all the discs to find the total volume of the solid.

3. What are the key components needed to use the disc method to find volume with trigonometric functions?

To use the disc method to find volume with trigonometric functions, you will need to know the limits of integration (the boundaries of the shape being rotated), the function being rotated, and the axis of rotation. You may also need to use trigonometric identities and integration techniques to simplify the calculation.

4. What are some common examples of using the disc method to find volume with trigonometric functions?

Some common examples of using the disc method to find volume with trigonometric functions include finding the volume of a cone, a sphere, or a paraboloid. These shapes can all be formed by rotating a two-dimensional shape (such as a circle, ellipse, or parabola) around an axis.

5. What are the advantages of using the disc method to find volume with trigonometric functions?

The disc method is a straightforward and efficient way to find the volume of a solid of revolution, as it only requires basic integration and trigonometry skills. It also allows for the calculation of volumes for more complex shapes that cannot be easily solved using other methods. Additionally, the disc method can be extended to find the volume of solids with varying densities, making it a useful tool in physics and engineering applications.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
728
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
978
Replies
1
Views
850
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
20
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
954
  • Calculus and Beyond Homework Help
Replies
34
Views
2K
  • Calculus and Beyond Homework Help
Replies
13
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
Back
Top