Calc 3 problems

[Sidoh]

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. :)

[Camel]

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?

[nslay]

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.

[Sidoh]

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.

[Camel]

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.