Recent content by shards5

Homework Statement Evaluate the triple integral \int\int\int^{}_{E} xy dV where E is the tetrahedron (0,0,0),(3,0,0),(0,5,0),(0,0,6). Is there a simple way to simplify the integration? Homework Equations The Attempt at a Solution \frac{z}{6} + \frac{y}{5} + \frac{x}{3} = 1 z =...

I guess I shouldn't have rushed through the calculations as I did, thanks a lot for pointing out my mistake.

Didn't even think about that . . .but for some reason it doesn't seem to be working. I switched the integration around and got the following. \int_0^1 \int_0^4 5ye^{-xy} dxdy After the first integration it comes out really neatly as -5e-xy and after plugging in 0 and 4 I get -5e-4y...

Homework Statement \int\int\int^{}_{B} ye^(-xy) dV where B is the box determined by 0 \leq x \leq 4, 0 \leq y \leq 1, 0 \leq z \leq 5.Homework Equations The Attempt at a Solution \int^{4}_{0}\int^{1}_{0}\int^{5}_{0} ye^(-xy) dzdydx Integrating the first time I get zye-xy Plugging in 5 and 0 I...

Okay, then after integration I get the following. 128*pi/2 - [1/2\theta + sin(2*\theta)/4] Plugging in my values I get the following. 128*pi/2 - pi/4 - sin(pi)/4 + sin(0)/4 but this is wrong. I am not sure what I am doing wrong.

That seems to make sense since if I think of \theta as how much an imaginary line sweeps through and radius as how long that line actually is. So now that I have the following: \int^{\pi/2}_{0}\int^{16}_{0} r drd\theta - \int^{\pi/2}_{0}\int^{16cos\theta}_{0} r drd\theta After the first...

I sketched it out and for the second one I am basing my intervals of integration of r from the center of the circle to the edge of the circle.That seems to make the most sense to me, however, a similar example in my book (which has a circle shifted 2 to the right instead of 8 and has a radius of...

Homework Statement Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant between the circles x2 + y2 = 256 and x2 - 16x + y2 = 0. Homework Equations The Attempt at a Solution Finding the intervals of integration for the polar coordinates. From the...

Oh I am so sorry about that, I completely mistyped that. It was supposed to be dx/du, dx/dv, dy/du, dy/dv. Also thanks you pointed out where I got it wrong. That helped a lot.

I think I calculated my Jacobian correctly since I am taking the correct partial derivatives of my functions. dx/du -1/7u + 3/7v = -1/7 dx/dv -1/7u + 3/7v = 3/7 dy/du 2/7u - 3/7v = 2/7 dy/dv 2/7u - 3/7v = 3/7 -1/7*2*7-3/7*3/7 = -2/49-9/49 = -11/49 and absolute value gives me 11/49

Homework Statement Suppose D is the parallelogram enclosed by the lines 2x-3y = 0, 2x-3y = 2, 3x-y = 0 and 3x-y = 1. \int\int^{}_{D} [(2x-3y) e^(3x-y) dA Homework Equations The Attempt at a Solution Set u to be equal to 2x-3y -> x = (u+3y)/2 Set v to be equal to 3x-y -> v = 3/2(u+3y)-y ->...

It was cos(x^2), and I think I got it using integration by substitution. Thanks a lot.

Homework Statement Evaluate the integral by reversing the order of integration. \int^{3}_{0}\int^{9}_{y^2} y cos(x^2) dydx Homework Equations ...? The Attempt at a Solution Drawing the picture out we get a sideways parabola. From the picture I get the following intervals of integration...

Homework Statement Find the volume of the solid bounded by the planes x = 0, y = 0, z = 0, and x + y + z = 9. Homework Equations . . . ? The Attempt at a Solution After drawing out the picture with z=0 I have a line going from 0,9 to 9,0 bounded by the x and y-axis giving me a triangle...

You, were right! I screwed up by substituting the limits into the y instead of x by mistake. Thanks a lot.