Recent content by tix24

how can i solve the integral with simple subsittution? if i break the integral up into 2 diferent integrals and use the sum rule, in one integral i have a square root and u substituion does not hold for this case as far as i can tell, so how would i go about solving this problem?

i don't know how to do the transformations; owever, i have come up with the integral, the problem is i don't know how to compute it. i came up with the integral in rectangular form and it goes as follows ∫dx∫dy∫dz where the limits of integration for x are 0 to 1/√2 for y are 0 to √(1-2x^2) for...

i have also figuired the following integral out but I am not sure if it correct. i did this one in rectangular coordinates. 4(∫dx∫dy∫dz) where the limits of integration for x are 0 to 1/√2 for y are 0 to √(1-2x^2)...

Homework Statement a) sketch the region in the first octant bounded by the elliptic cylinder 2x^2+y^2=1 and the plane y+z=1. b) find the volume of this solid by triple integration. Homework EquationsThe Attempt at a Solution I have already sketched the elliptic cylinder and the plane. my...

the equation provided fro the cone starts at (0,0,4) and expands downwards, the trace of the cone on the xy plane is a circle of radius 4. the sphere has a radius of 4. the cone is inside the sphere, and in the xy plane they both have the same radius. the cone is no longer inside the sphere...

did you mean to say it is outside of the cone? because the cone is only in the top half of the sphere, it leaves the sphere after the top half. (the radius exceeds that of a sphere.)

2pi just represents the constant in the equation, Ybar represents the distance from the axis which we are going to rotate the region at and A(D) is the area of the region. if i am going to find the volume of the bounded area then i have to find the area of this region. i think i can do this by...

fair enough, how do i use the theorem of pappus, can you aid me in this procedure? i know that the formula for volume using the theorem of pappus is as follows v=2pi(Ybar)A(D) but i am having trouble setting it up.

why shouldn't i include the lower half of the sphere?

im getting confused because the cone is only in the top half of the sphere, and in the bottom half of the sphere we have nothing. i don't know if the integral which i have set up is correct or not.

Homework Statement sketch the solid region contained within the sphere, x^2+y^2+z^2=16, and outside the cone, z=4-(x^2+y^2)^0.5. b) clearly identifying the limits of integration, (using spherical coordinates) set up the iterated triple integral which would give the volume bounded by the...

my my integral expression was as follows: integral dtheta (bounds from -pi/2 to pi/2) integral rdr (bounds from r=0 and r=cos(theta) and the expression which i integrate is x. which i wrote as x=rcos(theta)

So I set it up the following way: The integral for theta went from -pi/2 to pi/2 and the integral for r went from 0 to cos (theta) the function which I integrated was rcos (theta) the result was pi/8 Can anybody confirm this?

So since the circle is capped be the plane, we have to integrate over the function f (x,y)=x which in polar coordinates is rcos (theta). I'm still not a 100% sure if this is correct or not.

to calculate area by double integral i would have to integrate over the constant 1; my bounds of integration would be from r=0 to r=cos(theta) and i would have the second integral from theta=-Pi/2 to theta=Pi/2 for the theta integral