1. ## Surface Integral

Hey I'm stuck on this problem. I don't know if I'm doing something wrong with the surface integral or not because after simplification, I got to an integral that I can't do. Here is the question:

Evaluate the surface integral:

$\int\int_S yz dS$
S is the surface with parametric equations x = u^2, y = usinv, z = ucosv 0 <= u <= 1, 0 <= v <= $\pi/2$

This is attempt:

$\vec{r}_u = <2u, sinv, cosv>$

$\vec{r}_v = <0, ucosv, -usinv>$

$\vec{r}_u \times \vec{r}_v = <-u, 2u^2sinv, 2u^2cosv>$

$|\vec{r}_u \times \vec{r}_v| = u\sqrt{4u^2 + 1}$

$\int \int_D (usinv)(ucosv)|\vec{r}_u \times \vec{r}_v| dA$

$\int_0^1 u^3\sqrt{4u^2+1}du \int_0^{\pi/2} sinv cosvdv$

I don't know how to evaluate the integral $\int_0^1 u^3\sqrt{4u^2+1}du$ :\. if my steps are incorrect, I would be obliged if someone can tell me what I'm doing wrong and how to evaluate that integral.

2. Originally Posted by lilaziz1
Hey I'm stuck on this problem. I don't know if I'm doing something wrong with the surface integral or not because after simplification, I got to an integral that I can't do. Here is the question:

Evaluate the surface integral:

$\int\int_S yz dS$
S is the surface with parametric equations x = u^2, y = usinv, z = ucosv 0 <= u <= 1, 0 <= v <= $\pi/2$

This is attempt:

$\vec{r}_u = <2u, sinv, cosv>$

$\vec{r}_v = <0, ucosv, -usinv>$

$\vec{r}_u \times \vec{r}_v = <-u, 2u^2sinv, 2u^2cosv>$

$|\vec{r}_u \times \vec{r}_v| = u\sqrt{4u^2 + 1}$

$\int \int_D (usinv)(ucosv)|\vec{r}_u \times \vec{r}_v| dA$

$\int_0^1 u^3\sqrt{4u^2+1}du \int_0^{\pi/2} sinv cosvdv$

I don't know how to evaluate the integral $\int_0^1 u^3\sqrt{4u^2+1}du$ :\. if my steps are incorrect, I would be obliged if someone can tell me what I'm doing wrong and how to evaluate that integral.

let $\displaystyle t=4u^2+1 \iff u^2=\frac{t-1}{4} \implies 2udu=\frac{dt}{4}$

This gives

$\displaystyle \int_{1}^{5}\left(\frac{t-1}{4} \right)\sqrt{t}\left( \frac{dt}{8}\right)=\frac{1}{32}\int_{1}^{5}t^{\fr ac{3}{2}}-t^{\frac{1}{2}}dt$

3. ooh alright. I got another question. How do you look at an integral like that and be like "I have to do substitution here!" If I had an integral like $\int x\sqrt{4x^2 +1}dx$, I can see that I have to do substitution, but for the integral in the first post, I wouldn't know where to start.

4. Originally Posted by lilaziz1
ooh alright. I got another question. How do you look at an integral like that and be like "I have to do substitution here!" If I had an integral like $\int x\sqrt{4x^2 +1}dx$, I can see that I have to do substitution, but for the integral in the first post, I wouldn't know where to start.
What TheEmptySet did was see that " $4x^2+ 1$" in the square root and look for an "x" to put with the dx!

Instead of x outside the square root there is $x^3$ but TheEmptySet saw that that was $x^2(x)$ so he took out one x for the "x dx".

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# find the fundamental magnitudes for the surface x=ucosv,y=usinv,z=av

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