Recent content by lionely

I had got the answer from subtracting the section under the parabola from x =0 to x=3 from the cylinder formed by rotating the area under y=2. Thank you for all your responses

OKAY so what I did now was instead of really using calculus, I rotated the area bounded by y=2 from x=4 to x=0. That's a cylinder with radius 4 and height 2 so the volume would be 32Pi. Then from my previous work the volume bounded by the parabola x=3 and x=0 about the y-axis was 9pi/2...

Okay, I find that so weird, because the textbook I am using only has that method. So it should mean that method can be used. I will look up those methods though.. Also how would you even rotate an area under y=2 around the y axis. Since the integral would be something like ∫πx2 dy. There is no...

Homework Statement The area under y = (x^2/9) + 1 from x = 0 to x = 3 , and the area enclosed by the y= 0 , y=2 , x=3 , and x=4, are rotated about the y-axis , and the solid generated represents a metal ash tray , the units being cm. Calculate the volume of a metal. Homework EquationsThe...

Oh I got the answer now, I find it really hard to visualize these things... Thank you, also : my additional question, the only method I know is drawing a really thin rectangle of width dx or dy depending on which axis I am rotating it about. The question is , does the rectangle always have to...

I don't think so, when I rotated the area under y=4 from x=-1 to x=1 i got 32Pi ( This the area of the hole right? ) or is the area of the hole under the line y=1?

For the volume between y=4 and y=1 , the bounds I used were x= 1 and x=-1 and I got (32Pi) and added that to the 28Pi/3 . but that's not the answer. Still confused.

I have 16/x^4 but I wrote it badly and the volume of the middle region is that between the lines y=4 and y=1?

Homework Statement The area enclosed by y= 4/x^2 , y =1 and y =4 is rotated about the x-axis; find the volume generated. I am really confused I keep getting (14pi/3). But the answer in the back of the book is not that at all. An additional question, the only method I know is drawing a really...

Ohhhhhhhh. Oh okay

I just tried it over when I integrate the tangent at A from x=6 to x =3 I get 27 and when I integrate the tangent at O from x=3 to x= 0 I also get 27. So I see the 54m^2. But I still don't fully understand why if I use the tangent at A and use the whole interval, I get 108m^2. Do you have...

I integrated the equation of the tangent at A ; y = 36 -6x from x = 6 to x = 0. Also I integrated the equation of the tangent at O ; y = 6x for the same interval. Got 108m^2 in both cases. I was doing this because I thought the areas under either one of these curves would be the area of the...

Oh okay sorry. Area bounded the curve I'll call it C and area of Triangle AOT. C+AOT = 36 + 108 = 144 and AOT = 3C 108 = 3(36).. Hmmm so what should I do now?

Hold on I actually have three areas there. Should I ignore the 54m^2?

Homework Statement The parabola y = 6x - x^2 meets the x-axis at O and A. The tangent at O and A meet at T. Show that the curve divides the area of the triangle OAT into two parts in the ratio 2:1. Homework EquationsThe Attempt at a Solution Here is my working with my sketch. So I thought I...