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 =...
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...
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...
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...
Homework Statement
Suppose D is the parallelogram in the xy-plane with vertices P(-1,5), Q(1,-5), R(5,-1), S(3,9)
\int\int ^{}_{D} (6x+12y) dA
HINT: Use transformation x = \frac{1}{6}(u+v) and y = \frac{1}{6} (-5u+v).
Homework Equations
The Attempt at a Solution
Calculating the Jacobian I...
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...