I just got out of my final for Calculus 3 and they had a few interesting problems on there. I'm still not sure that I did them correctly, but I did end up getting complete answers for them.
Here they are.
1) Evaluate [tex]\int_{0}^{1}{\int_{y}^{1}{\sin(x^2)\,dx}\,dy}[/tex]
2) Find the point on the plane given by [tex]ax+by+cz=d[/tex] closest to the origin, then compute its distance.
There were a few less interesting ones involving Green's and Stoke's theorems. The problems were run of the mill -- the kind where they try to trick you into evaluating it as a multi-pieced curve integral. I'm sure everyone interested has seen enough of those, though. :)
Quote from: Sidoh on December 12, 2007, 11:05:52 AM
1) Evaluate [tex]\displaystyle\int_{0}^{1}{\int_{y}^{1}{\sin(x^2)\,dx}\,dy}[/tex]
It has been a while...
IIRC, [tex]\displaystyle\int\sin x dx = -\cos x[/tex], which when combined with the substitution rule...
u = x^2
du = 2dx -> dx = 1/2 du
[tex]\displaystyle\int_{y}^{1}{\sin\,x^2\,dx}[/tex]
[tex]\displaystyle\int_{y^2}^{1^2}{\sin\,u\,\frac{1}{2}du}[/tex]
[tex]\displaystyle\frac{1}{2}\int_{y^2}^{1}{\sin\,u\,du}[/tex]
[tex]\displaystyle\frac{1}{2}\left[-cos\,u\right]_{y^2}^{1}[/tex]
[tex]\displaystyle\frac{1}{2}\left[-\cos\,1+\cos\,y^2\right][/tex]
Finally,
[tex]\displaystyle\int_{0}^{1}{\frac{1}{2}\left[-\cos\,1+\cos\,y^2\right]\,dy}[/tex]
[tex]\displaystyle\frac{1}{2}\int_{0}^{1}{\cos\,y^2\,dy}-\frac{1}{2}\int_{0}^{1}{\cos\,1\,dy}[/tex]
[tex]\displaystyle\frac{1}{2}\int_{0}^{1}{\cos\,y^2\,dy}-\frac{1}{2}\cos\,1[/tex]
u = y^2
du = 2dy -> dy = 1/2 du
[tex]\displaystyle\frac{1}{2}\int_{0^2}^{1^2}{\cos\,u\,\frac{1}{2}du}-\frac{1}{2}\cos\,1[/tex]
[tex]\displaystyle\frac{1}{4}\int_{0}^{1}{\cos\,u\,du}-\frac{1}{2}\cos\.1[/tex]
[tex]\displaystyle\frac{1}{4}\left[\sin\,u\right]_{0}^{1}-\frac{1}{2}\cos\,1[/tex]
[tex]\displaystyle\frac{1}{4}\sin\,1-\frac{1}{2}\cos\,1[/tex]
amirite?
You made a small mistake in the beginning, which unfortunately propagated throughout the problem
[tex]
u = x^2 \quad
du = 2x dx
[/tex]
I wish this forum software supported the AMS TeX packages...I could not use the align environment.
The solution to that one is actually really simple. All you have to do is switch the order of integration. I immediately thought of that during the test, but somehow made an error in my mental calculations to see if it would work and decided that it wouldn't. Instead, I applied a Jacobian transformation which made the problem solvable. I think I made an algebraic error somewhere in my solution, but I ended up with the same form of answer that you get when you use the intended method.
nslay, I can take a look at that. I thought those packages were included, but I'll make sure they aren't and get them in there if they are not. Thanks for pointing it out.
Quote from: nslay on December 15, 2007, 12:09:43 PM
You made a small mistake in the beginning, which unfortunately propagated throughout the problem
[tex]
u = x^2 \quad
du = 2x dx
[/tex]
I wish this forum software supported the AMS TeX packages...I could not use the align environment.
d'oh!
I made the same mistake in the second substitution as well.