Recent content by Yami

Okay so if it's just {y > 0, but no x > 0, z > 0} OR {x > 0, but no y > 0, z > 0}, two quadrants of the space, then would it be ##0 < \theta < \pi##?

Alright, I got it. Thanks for your help. I think I get the gist of the radians restriction; so if we have just x > 0, y > 0, but no z > 0 would it be ##R = \{(r, \theta); 0 < r < \sqrt{3}, 0 < \theta < \pi\}##?

Homework Statement Show that ##\iint_{S}(x^2 + y^2)d\sigma = \frac{9\pi}{4}## where ##S = \{(x,y,z): x > 0, y > 0, 3 > z > 0, z^2 = 3(x^2 + y^2)\}## Homework Equations ##\iint_{S}f(x,y,z)d\sigma = \iint_{R}f(r(x,y))\sqrt{[r_x(x,y)]^2 + [r_y(x,y)]^2 + 1}## where ##r : R → ℝ^3, R \in ℝ^2##...

I just realized p and q are in ##ℝ^n##. So [p, q] isn't an interval I can integrate over. I just found the equation I should probably use: ##\gamma:[0,1]→ℝ^n## defined as ##\gamma (t) = tq + 1(1 - t)p## for ##t \in [0, 1]##. Thanks for the hint.

Homework Statement For two points p and q in ℝ^n, use the formula (20.3) to check that the arclength of the parametrized segment from p to q is ||p - q||. Homework Equations Formula (20.3): A smooth parametrized path \gamma: [a, b]→ℝ^nis rectifiable, and its arclength l is given by l...

I'm still confused... so none of what I did above is correct? Okay so you're saying this would work? \int\int_{D}|xyz|dxdydz=8\iiint_{D_{1}}xyzdxdydz D=\{ (x,y,z)|x^2+y^2+z^2 ≤ 1\} and D_{1}=\{ (x,y,z)|x^2+y^2+z^2 ≤ 1, x ≥ 0, y≥0, z ≥ 0\}

Okay I'm a little confused. I removed the abc from the integral for the sake of simplicity and tried plugging it into Matlab's triplequad() function. The answer is 1/6 according to Matlab, but obviously I need to know how to evaluate this by hand. So I'm trying to see how it got to 1/6...

Okay I went over it again and found a mistake so the new answer is: \frac{21\sqrt{2} - 18}{40\sqrt{2}} I also tried your suggestion of integrating dx first and set this up: \int_{0}^{1}\int_{\sqrt{0.5y}}^{\sqrt{y}}(x+y) dx dy which led to a different answer: \frac{21\sqrt{2} -...

Okay so I can't ignore the absolute value... And you're saying these are the wrong regions? \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{1}?

Homework Statement Evaluate \iiint_\textrm{V} |xyz|dxdydz where V = \{(x,y,z) \in ℝ^3:\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} ≤ 1\}Homework Equations Change of Variables Theorem: \int_\textrm{ψ(u)} f(x)dx = \int_\textrm{K} f(\Psi(u))|detD\Psi(u)|du Examples: 1) For a ball of...

I've got this: \frac{45\sqrt{2} - 32}{80\sqrt{2}} which is about 0.5. It feels right I guess. Thank you for your help.

Okay my next guess is this: \iint_\textrm{I}f = \int_{0}^{\frac{1}{\sqrt{2}}}\int_{x^2}^{2x^2}(x + y)dydx + \int_{\frac{1}{\sqrt{2}}}^{1}\int_{x^2}^{1}(x + y)dydx

I just realized I've been writing h(x) = x instead of x^2 as the lower bound. That's not the main reason it's wrong though, yes?

I did Okay, I would guess that it's \iint_\textrm{I}f = \int_{0}^{\frac{1}{\sqrt{2}}}\int_{x^2}^{2x^2}(x + y)dydx Am I warm?

Integral calculus involving Fubini's Theorem Homework Statement f(x,y) = x + y, if: x^2 ≤ y ≤ 2x^2 f(x,y) = 0, otherwise Evaluate \iint_\textrm{I}f where I = [0,1] x [0,1] Homework Equations For a Jordan domain K in ℝ^n, let h: K → ℝ and g: K → ℝ be continuous...