There are two problems I got stuck...
1. Is the area in the first quadrant bounded between the x-axis and the curve
\ y= \frac{x} {2*(x^2+2)^{7/8}}
finite? This one, I used the Area formula... but then I cannot integrate it... and then how to determine if it's finite or not?
2...
[FONT=Arial]The homework question is this:
Prove If D is a bounded subset of R then D bar = D U D’ is also bounded where D’ is the set of accumulation points of D.
What is a general outline of a proof?
Here is the problem:
First Part (already done): Find the volume of the solid that is bounded above by the cylinder z = 4 - x^2, on the sides by the cylinder x^2 + y^2 = 4, and below by the xy-plane.
Answer: \int_{-2}^{2}\int_{-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}}\int_{0}^{4 -...
Here is the problem:
Find the volume of the solid that is bounded above by the cylinder z = 4 - x^2, on the sides by the cylinder x^2 + y^2 = 4, and below by the xy-plane.
Here is what I have:
\int_{-2}^{2}\int_{-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}}\int_{0}^{4 - x^2}\;dz\;dy\;dx\;=\;12\pi...
Evaluate the integral over the bounded closed region B
\int \int_{B} x^2 y^3 dx dy B is the region bounded by y=x^2 and y =x
certainly the x=0 to x=1 will be the limits for the x part
but what about the y does it go from root y to y?
please help! i neeed to understand how to setup...
Find the area of the region bounded by: r= 6-2sin(\theta)
here's what i did:
6-2sin(\theta) = 0
sin(\theta) = 1/3
so the bounds are from arcsin(-1/3) to arcsin(1/3) right?
my integral:
\int_{-.339}^{.339} 1/2*(6-2sin(\theta))^2
i get a answer of 0.6851040673*10^11, and it's...
How do I do this problem below? Plz guide me step by step or a least show me the correct bounds so I can learn...thanks
***Find the area bounded by y^2=x and y=x=2
Thanks so much
problem:
find volume bordered by cylinder x^2 + y^2 = 4 and y+z=4 and z=4.
the answer is said to be 16p. but I couldn't find it.
I found it in double integral part.so it must be solved with double integral. I tried with Jacobian tranformation. nut still couldn't solve it. I was confused...
Find the area bounded by the two curves:
x=100000(5*sqrt(y)-1)
x=100000(\frac{(5*sqrt(y)-1)}{(4*sqrt(y))})
i'm having a lot of trouble trying to find the lower and upper limit of the two functions. I tried setting the two functions together and solving for y, but i get 0. then trying to...
could someone give me an example of a function that is bounded but is nonintegrable?
i need to know what a nonintegrable function bounded on [a,b] is as said in my preperation file for a test? urgent help needed
If A is a bounded subset of the reals, show that the points infA, supA belong to the closure A*.
At first the answer seems obvious to me since A* contains its limit points. I'm just having trouble putting it into words, any suggestions would be great, thanks.
This is the problem: find the area of the region bounded by the curves f(x) = x^2 + 2 and g(x) = 4 - x^2 on the interval [-2,2]
I did the whole integral from -2 to 2 with (4-x^2) - (x^2 + 2) dx because the graph of g(x) is on top between the region bounded. But from my drawing, the points...
Can someone please help me w/ these problems below:
Find the volume generated by revolving the area abounded by x=y^2 and x=4 about
a)the line y=2
b)the line x= -1
***I tried to write out the integral not sure if it's correct:
a) V=pi* int. of (sqrt(x)^2-(2-sqrt(x))^2) dx
**integral form...
I don't know if anyone will be able to help me, I am really stuck on this question!
"Show that the volume of the region bounded by the cone
z=sqrt((x*x)+(y*y)) and the parabloid z=(x*x)+(y*y) is
PI/6"
The bits in the brackets (ie x*x and y*y) are x squared and y squared respectively and...