Calculus 2 volume integration, y=axis

In summary, the person is having trouble with a math problem and is confused on how to rotate a bounded region around the y-axis. They are looking for help from someone else or for direction on how to rotate the bounded region.
  • #1
funkyguy4000
3
0
Hello, so I've been having lots of trouble with a math problem.
I can differentiate functions but I'm kinda confused on what to do when it is like y=4

So say you have the given: y= x^2 and then y=4, x=0, x=2. Rotate about the y-axis.
I'm know you have to get it into respect to y so x=sqrt(y) but what would I do from there?! you can't put x in terms of y for y=4! I'm confused. Can somebody help me out or point me in the right direction.
 
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  • #2
they are giving you 4 different curves or lines and you are supposed to envision the region bounded by those curves or lines. so you have the line y=4 at the top, y=x^2 on the right, ...

ooops, your equations do not fit together to give one region. you need to re - read the question.
 
  • #3
Okay well the y=x^2 is a parabola so it goes on both sides of the y-axis. And then the y=4 is another bounding line.

we are supposed to rotate the bounded region from x=0 to 2 around the y-axis.

I'm just confused on how to rotate it around like that.
 
  • #4
well you don't want to say x=2 if you know y=4 and y = x^2. rather you should say y = x^2 for x≥0, y=4, and x=0. but if that's what it said, its their fault not yours.

so apparently the region is bounded by y = x^2 for x≥0, x = 0 and y=4. that makes it a portion of the right half of the parabola. then you rotate it around the y axis. is that hard to see? your book may not be too well written.
 
  • #5
well they give me the limits x=0 and x=2. y=x^2 and y=4 intersect at x=2.

I can imagine it. I'm just confused how to rotate it around the y=axis. what method to use and how to get the variables needed in the method. I tried the shells method and that didn't work. I didn't understand how to get the needed material.
 

FAQ: Calculus 2 volume integration, y=axis

What is volume integration in Calculus 2?

Volume integration in Calculus 2 is a method of finding the volume of a three-dimensional shape using the concept of integration. It involves dividing the shape into infinitely small slices, finding the area of each slice, and then adding up all the areas to find the total volume.

What is the y-axis in Calculus 2 volume integration?

The y-axis is one of the axes in a 3D coordinate system. It is perpendicular to the x-axis and the z-axis, and it is used to represent the height or depth of a shape in volume integration. In y-axis volume integration, the slices are taken parallel to the y-axis to calculate the volume.

How is volume integration different from area integration?

Volume integration is used to find the volume of a 3D shape, while area integration is used to find the area of a 2D shape. In volume integration, the slices are taken perpendicular to the axis of rotation, while in area integration, the slices are taken parallel to the axis of rotation.

What is the formula for calculating volume using y-axis integration?

The formula for calculating volume using y-axis integration is V = ∫a^b A(y) dy, where a and b are the limits of integration, A(y) is the area of the slice at a given y-value, and dy represents the infinitely small change in the y-value.

What are some real-life applications of volume integration in Calculus 2?

Volume integration is used in various fields, such as engineering, physics, and architecture, to calculate the volume of 3D objects. It is also used in fluid mechanics to calculate the volume of fluids in tanks or pipes. Additionally, volume integration is used in computer graphics to create 3D models and in medical imaging to calculate the volume of organs or tumors.

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