Calculus Help: Find Volumes & Solve Problems

In summary, the person is asking for help with calculus problems. They say that they got stuck at one point and need help getting started. They list the problems that they are having and ask for help with them.
  • #1
gigi9
40
0
Calculus help please!

Can someone please help me to figure out how to do these problems below?...Or at least get me started in the right direction. Thanks a lot for ur help!I would really appreciate ur help!

1) Find the volume of the solid of revolution generated when the area bounded by the given curve is revolved about the x-axis.
x^(2/3)+y^(2/3)=a^(2/3), first quadrant.
****I got to this point: Integral from 0 to a of the
pi*(-x^2/3+a^2/3)^3 dx *** Is this right? what should I do next if it's right?

2)The square bounded by the axes and the lines x=2, y=2 is cut into two parts by the curve y^2=2x. These two are areas are revolved about the line x=2. Find the volume generated.
*** for Volume 1, I have integral from 0 to 2 of pi*[2-(y^2)/2]^2 dy = pi*[4x-2/3y^3+1/20y^5] from 0 to 2 = 64*pi/5 <--ans

*** for Volume 2, I used integral from 0 to 2 of pi*[(y^2)/2]^2 dy = pi* 1/20x^5 from 0 to 2 = 8/5*pi <--ans
Did I do these two problems correctly? If not or any thing wrong, tell me how to fix it... Thanks.

3) If the area bounded by the parabola y=H-(H/R^2)x^2 and the x-axis is revolved about the y-axis, the resulting bullet-shaped solid is a segment of a paraboloid of revolution w/ height H and radius of base R. Show that its volume is half of the volume of the circumscribing cylinder. ****Plz tell me how to get started to this problem, if you can post a graph on here, please show me the graph also. Thank you very very much for ur help!
 
Physics news on Phys.org
  • #2
For problem 1, what you have is correct. What you need to do next is to do the integration! It's probably simplest just to go ahead and multiply (a^(2/3)- x^(2/3))^3.

For problem 2, volume 1, the area is rotated around the line x= 2. The "washer" created by rotating the region between x= 0 and x= y^2/2 has area &pi;(r12- r22) where r1 is the distance from x= 2 to 0 (i.e. 2) and r2[/sup] is the distance from x= 2 to x= y^2/2 (i.e. y^2/2- 2). Your integral should be &pi;&int;02((y^2/2-2)2-4)dy= &pi;&int;02(y4/4- 2y2)dy

For volume 2, you have a disk with radius 2-x= 2- y^2/2.

The formula you are using appears to be rotating around the y-axis.

For problem 3, you certainly ought to be able to graph y= H-(H/R^2)x^2 yourself. That is a parabola with vertex at (0,H) that crosses the x-axis at (R,0) and (-R,0). Of course, since you are rotating it around the y-axis, you only need the portion in the first quadrant.

Each "disk" will have area &pi;x2= &pi;(R2/H)(H-y) (solve y= H- (H/R2)x2 for x2).
The integral for the volume is &int(R2/H);0H(H-y)dy.

Once you have found that, divide by the volume of a cylinder of height H and radius R to find the fraction.
 
  • #3
For problem 2, volume 1, the area is rotated around the line x= 2. The "washer" created by rotating the region between x= 0 and x= y^2/2 has area &pi;(r12- r22) where r1 is the distance from x= 2 to 0 (i.e. 2) and r2[/sup] is the distance from x= 2 to x= y^2/2 (i.e. y^2/2- 2). Your integral should be &pi;&int;02((y^2/2-2)2-4)dy= &pi;&int;02(y4/4- 2y2)dy

For volume 2, you have a disk with radius 2-x= 2- y^2/2.

The formula you are using appears to be rotating around the y-axis.

Can you explain more on this one please, where did u get ƒÎç02((y^2/2-2 )2-4)dy ? Where does the "2" inside the () come from?
 

1. What is calculus and why is it important?

Calculus is a branch of mathematics that deals with the study of change and motion. It is important because it helps us understand and model real-world phenomena, such as the motion of objects, the growth of populations, and the change in temperature over time.

2. How do you find the volume of a three-dimensional shape using calculus?

To find the volume of a three-dimensional shape using calculus, you first need to set up an integral that represents the volume of the shape. This integral will involve the area of cross-sections of the shape, which can be found using integrals. Then, evaluate the integral to find the volume.

3. Can calculus be used to solve real-world problems?

Yes, calculus can be used to solve real-world problems. For example, it can be used to optimize the production of a certain product, determine the maximum capacity of a container, or find the shortest path for a delivery truck. Calculus is a powerful tool for modeling and solving many real-world problems.

4. What are some common applications of calculus?

Calculus has many applications in various fields such as physics, engineering, economics, and biology. Some common applications include finding the area and volume of complex shapes, optimizing functions to find maximum or minimum values, and determining rates of change in natural phenomena.

5. How can I improve my understanding and skills in calculus?

To improve your understanding and skills in calculus, it is important to practice solving problems and to seek help when needed. You can also review the fundamental concepts and formulas, and try to apply them to real-world scenarios. Additionally, online resources and textbooks can be helpful for additional practice and learning.

Similar threads

Replies
4
Views
191
  • Calculus
Replies
16
Views
352
  • Calculus
Replies
3
Views
1K
  • Calculus
Replies
2
Views
1K
Replies
20
Views
2K
Replies
1
Views
813
Replies
3
Views
533
Replies
1
Views
1K
Replies
2
Views
801
Replies
2
Views
1K
Back
Top