Calculus Problems

[ryuen]

these problems have been makin me mad so anyone have a step by step solution or just solution?

1. find area between curve

x+y = 0
x+ y^2 =30


2. Find area bounded by given curves rotating around axis.

y=9x^2, x=1, y=0, around x axis

3. same as above

y=x^2, y=1, x=0, x=1, around y axis

[Schalldampfer]

these problems have been makin me mad so anyone have a step by step solution or just solution?

1. find area between curve

x+y = 0
x+ y^2 =30


2. Find area bounded by given curves rotating around axis.

y=9x^2, x=1, y=0, around x axis

3. same as above

y=x^2, y=1, x=0, x=1, around y axis

Heh. This takes me back a year...
For number 1, find the two intersection poitns of the curves.
Now, integrate the one that is above the other from the lower intersection point to the higher one. Now, integrate the bottom one from the same point to the other point. Subtract, and voila! There you have it.

[Schalldampfer]

Now, then, for numbers 2 and 3... holy crap it's hard to explain online. But basically, what you have to do is use the washer method or the shell method. Try searching for them online. It's rather hard to explain without pictures. If you still have further questions, please reply.

[ryuen]

well i know that. but this problem is weiirrrdd. take a look at it. washer and shell i know the methods too but i cant do it. thats why i need someone to solve it out if possible

[Frodo Baggins]

these problems have been makin me mad so anyone have a step by step solution or just solution?

1. find area between curve

x+y = 0
x+ y^2 =30


2. Find area bounded by given curves rotating around axis.

y=9x^2, x=1, y=0, around x axis

3. same as above

y=x^2, y=1, x=0, x=1, around y axis

I'm not sure about any of this. It's been a long time and I don't have a text on hand so I basically derived all the formulas here.

y = 9x^2, rotated about the x-axis between x = 0 and x = 1

The area element here are disks with area pi(9x^2)^2 = 81pix^4
Thus, the volume is given by the integral of (81 pi x^4 dx) between x = 0 and x = 1

See if you can do #3 on your own. You have a lot of freedom as to how to do it. You can do it the same way as above (disk/washer method) by using x = sqrt(y) (you only need the positive branch), or you can do it via shell method.

Actually lemme try it.

The function is x = sqrt(y) from y = 0 to y = 1 (why is it y = 1?)

Again, the area elements are disks, Pi(sqrt(y))^2 = (pi)y

Thus, the integral yielding the volume is V = S[ (pi)y dy, y = 0...1]

Hope that helped (and it would help if someone double checked...it's been a long time since I've touched this type of Calculus). But I've got to jet to my class.