The problem I am to solve goes like this:
Find the area of the cap cut from the paraboloid y^2 +z^2=3x by the plane x=1.
Here is what I have so far;
\(\displaystyle \[
x = \frac{{y^2 + z^2 }}
{3} \to parametric\;equation \to s(u,v) = u^2 i\limits^ \wedge + \sqrt 3 u\sin vj\limits^ \wedge + \sqrt 3 u\cos vk\limits^ \wedge
\]\)
I am not sure if I put the sine and cosine in the right place. I assume the placement of the sine and cosine doesn't matter because cos^2+sin^2=1, no matter the orientation. Am I correct?
For my limits on my parametric equation I have
\(\displaystyle \[
0 \le u \le \sqrt 3 \;and\;0 \le v \le 2\pi
\]\)
Assuming my limits are correct the next thing I did was find the partial of my parametric equation (which I will call from now on r) with respect to first u then v.
\(\displaystyle \[
r_u = 2ui\limits^ \wedge + \sqrt 3 \sin (v)j\limits^ \wedge + \sqrt 3 \cos (v)k\limits^ \wedge \hfill \\
r_v = 0i\limits^ \wedge + \sqrt 3 *u*\cos (v)j\limits^ \wedge + \sqrt 3 *u*\sin (v)k\limits^ \wedge \hfill \\
\]\)
Next I cross the two vectors above obtaining;
\(\displaystyle \[
r_u \;x\;r_v = \left\langle {u*(3 - 6*(\cos (v))^2 , - 2*\sqrt 3 *u^2 *\sin (v),2*\sqrt 3 *u^2 *\cos (v)} \right\rangle
\]\)
The next thing I did was obtain the magnitude of the above vector with the following results.
\(\displaystyle \;\left\| {r_u \;x\;r_v } \right\| = \sqrt {(u*(3 - 6*(\cos (v))^2 )^2 + ( - 2*\sqrt 3 *u^2 *\sin (v))^2 + (2*\sqrt 3 *u^2 *\cos (v))^2 }\)
Lastly I put together the double integral using the formula in my text for Area of a parametric surface. Is my integral set up properly? Here she is. :roll:
\(\displaystyle \int\!\!\!\int\limits_s {ds} = \int\!\!\!\int\limits_D {\;\left\| {r_u \;x\;r_v } \right\|} \;dA = \int\limits_0^{2\pi } {\int_0^{\sqrt 3 } {\sqrt {(u*(3 - 6*(\cos (v))^2 )^2 + ( - 2*\sqrt 3 *u^2 *\sin (v))^2 + (2*\sqrt 3 *u^2 *\cos (v))^2 } } } \;du\;dv\) And Lastly, assuming I did all this correct, could I get some tips on simplifying the integrand and actually integrating it? Is there a way to simplify it to a point where I could get my TI89 or Mathematica or Maple (all of which I have access to here on my computer) to evaluate it, without giving me an error message?
\(\displaystyle {
& \;\left\| {r_u \;x\;r_v } \right\| = \sqrt {9u - (36u*\sin (v)^2 \cos (v)^2 ) + 12u^4 *\sin (v)^2 + 12u^4 \cos (v)^2 } \Rightarrow \cr
& \sqrt {9u(1 - (4*\sin (v)^2 \cos (v)^2 )) + 12u^4 *(\sin (v)^2 + \cos (v)^2 )} \Rightarrow \cr
& \sqrt {9u(1 - (\sin (2v))^2) + 12u^4 } \cr}\)
Am I correct? I think I am, but then again when I put it into my TI89 and hit enter it takes it about 15 seconds to spit out the same double integral with everything the same, but the integrand reverts back to one of the colapsed forms I have in the beginning...or similar to it anyway. I am getting frustrated. I don't see anything that could help me in the table. At least not in this form. Hopefully I didn't mess it up.
Addendum: I finally got an answer after taking the integral that my 89 spit out and putting it in Mathematica!! Also I realized that I forgot to modify my u limits of integration from 0 to sprt[3] to 0 to 1. Here is the result for those interested.
\(\displaystyle \sqrt 3 \;\int\limits_0^{2\pi } {\int\limits_0^1 {\sqrt {3*Cos(2v)^2 4u^3 *u} } \,du\,dv = 8}\)
Find the area of the cap cut from the paraboloid y^2 +z^2=3x by the plane x=1.
Here is what I have so far;
\(\displaystyle \[
x = \frac{{y^2 + z^2 }}
{3} \to parametric\;equation \to s(u,v) = u^2 i\limits^ \wedge + \sqrt 3 u\sin vj\limits^ \wedge + \sqrt 3 u\cos vk\limits^ \wedge
\]\)
I am not sure if I put the sine and cosine in the right place. I assume the placement of the sine and cosine doesn't matter because cos^2+sin^2=1, no matter the orientation. Am I correct?
For my limits on my parametric equation I have
\(\displaystyle \[
0 \le u \le \sqrt 3 \;and\;0 \le v \le 2\pi
\]\)
Assuming my limits are correct the next thing I did was find the partial of my parametric equation (which I will call from now on r) with respect to first u then v.
\(\displaystyle \[
r_u = 2ui\limits^ \wedge + \sqrt 3 \sin (v)j\limits^ \wedge + \sqrt 3 \cos (v)k\limits^ \wedge \hfill \\
r_v = 0i\limits^ \wedge + \sqrt 3 *u*\cos (v)j\limits^ \wedge + \sqrt 3 *u*\sin (v)k\limits^ \wedge \hfill \\
\]\)
Next I cross the two vectors above obtaining;
\(\displaystyle \[
r_u \;x\;r_v = \left\langle {u*(3 - 6*(\cos (v))^2 , - 2*\sqrt 3 *u^2 *\sin (v),2*\sqrt 3 *u^2 *\cos (v)} \right\rangle
\]\)
The next thing I did was obtain the magnitude of the above vector with the following results.
\(\displaystyle \;\left\| {r_u \;x\;r_v } \right\| = \sqrt {(u*(3 - 6*(\cos (v))^2 )^2 + ( - 2*\sqrt 3 *u^2 *\sin (v))^2 + (2*\sqrt 3 *u^2 *\cos (v))^2 }\)
Lastly I put together the double integral using the formula in my text for Area of a parametric surface. Is my integral set up properly? Here she is. :roll:
\(\displaystyle \int\!\!\!\int\limits_s {ds} = \int\!\!\!\int\limits_D {\;\left\| {r_u \;x\;r_v } \right\|} \;dA = \int\limits_0^{2\pi } {\int_0^{\sqrt 3 } {\sqrt {(u*(3 - 6*(\cos (v))^2 )^2 + ( - 2*\sqrt 3 *u^2 *\sin (v))^2 + (2*\sqrt 3 *u^2 *\cos (v))^2 } } } \;du\;dv\) And Lastly, assuming I did all this correct, could I get some tips on simplifying the integrand and actually integrating it? Is there a way to simplify it to a point where I could get my TI89 or Mathematica or Maple (all of which I have access to here on my computer) to evaluate it, without giving me an error message?
\(\displaystyle {
& \;\left\| {r_u \;x\;r_v } \right\| = \sqrt {9u - (36u*\sin (v)^2 \cos (v)^2 ) + 12u^4 *\sin (v)^2 + 12u^4 \cos (v)^2 } \Rightarrow \cr
& \sqrt {9u(1 - (4*\sin (v)^2 \cos (v)^2 )) + 12u^4 *(\sin (v)^2 + \cos (v)^2 )} \Rightarrow \cr
& \sqrt {9u(1 - (\sin (2v))^2) + 12u^4 } \cr}\)
Am I correct? I think I am, but then again when I put it into my TI89 and hit enter it takes it about 15 seconds to spit out the same double integral with everything the same, but the integrand reverts back to one of the colapsed forms I have in the beginning...or similar to it anyway. I am getting frustrated. I don't see anything that could help me in the table. At least not in this form. Hopefully I didn't mess it up.
Addendum: I finally got an answer after taking the integral that my 89 spit out and putting it in Mathematica!! Also I realized that I forgot to modify my u limits of integration from 0 to sprt[3] to 0 to 1. Here is the result for those interested.
\(\displaystyle \sqrt 3 \;\int\limits_0^{2\pi } {\int\limits_0^1 {\sqrt {3*Cos(2v)^2 4u^3 *u} } \,du\,dv = 8}\)