Multiple, Line and Surface Integrals: Problems with Solutions

Solution:
The region [tex]D[/tex] is a rectangle and the integrand is continuous, so the double integral equals an iterated integral. We integrate first with respect to [tex]y[/tex]; the term [tex]\sin (\pi x)[/tex] does not depend on [tex]y[/tex] and the [tex]y[/tex]-interval has length [tex]1[/tex]:
[tex]\int_{2}^{3}\left(x^2y+\sin (\pi x)\right)dy=x^2\times \frac{3^2-2^2}{2}+\sin (\pi x)=\frac{5}{2}x^2+\sin (\pi x)[/tex]
Then
[tex]\int_{0}^{1}\left(\frac{5}{2}x^2+\sin (\pi x)\right)dx=\frac{5}{6}+\left[-\frac{\cos(\pi x)}{\pi}\right]_{0}^{1}=\frac{5}{6}+\frac{1}{\pi}+\frac{1}{\pi}=\frac{5}{6}+\frac{2}{\pi}[/tex]
The sine term does not vanish: [tex]\sin (\pi x)\ge 0[/tex] on [tex][0, 1][/tex], so it adds the positive amount [tex]\frac{2}{\pi}[/tex].

Solution:
The three lines bound the triangle with vertices [tex](0, 0)[/tex], [tex](0, \pi)[/tex] and [tex](\pi, \pi)[/tex], so
[tex]D=\{(x,y):\ 0\le y\le \pi,\ 0\le x\le y\}[/tex]
The integrand is continuous, so the double integral equals the iterated integral. Integrating first with respect to [tex]x[/tex]:
[tex]\int_0^{y}\cos(x+y)\,dx=\sin (x+y)\Big|_{x=0}^{x=y}=\sin 2y-\sin y[/tex]
Hence
[tex]\iint_D\cos(x+y)\,dx\,dy=\int_0^{\pi}(\sin 2y-\sin y)\,dy=\left(-\frac{\cos 2y}{2}+\cos y\right)\Big|_0^{\pi}=\left(-\frac{1}{2}-1\right)-\left(-\frac{1}{2}+1\right)=-2[/tex]

Solution:
Intersecting consecutive lines gives the vertices: [tex]y=1[/tex] and [tex]y=3x+1[/tex] meet at [tex](0, 1)[/tex]; [tex]y=3x+1[/tex] and [tex]x+y=5[/tex] give [tex]4x=4[/tex], i.e. [tex](1, 4)[/tex]; [tex]x+y=5[/tex] and [tex]x=2[/tex] meet at [tex](2, 3)[/tex]; [tex]x=2[/tex] and [tex]y=1[/tex] meet at [tex](2, 1)[/tex].
The lower boundary is [tex]y=1[/tex] everywhere, while the upper boundary is [tex]y=3x+1[/tex] for [tex]0\le x\le 1[/tex] and [tex]y=5-x[/tex] for [tex]1\le x\le 2[/tex], because [tex]3x+1\le 5-x\iff x\le 1[/tex]. So we split the region by the line [tex]x=1[/tex]:
[tex]S=\iint_D dx\,dy=\int_{0}^{1}\int_{1}^{3x+1}dy\,dx+\int_{1}^{2}\int_{1}^{5-x}dy\,dx=\int_{0}^{1}3x\,dx+\int_{1}^{2}(4-x)\,dx=\frac{3}{2}+\frac{5}{2}=4[/tex]

Solution:
The direction vector is [tex]\overrightarrow{AB}=(1, -1, 5)[/tex], so the segment is parametrized as follows, where [tex]t=0[/tex] gives [tex]A[/tex] and [tex]t=1[/tex] gives [tex]B[/tex] (for an integral of the second kind the orientation matters):
[tex]x=1+t,\quad y=2-t,\quad z=-1+5t,\quad 0\le t\le 1[/tex]
Then [tex]dx=dt[/tex], [tex]dy=-dt[/tex], [tex]dz=5\,dt[/tex] and
[tex]x+y+1=4,\quad 2y-3=1-2t,\quad 3x+2y-z=(3+3t)+(4-2t)-(-1+5t)=8-4t[/tex]
Hence the integral equals
[tex]\int_{0}^{1}\left(4-(1-2t)+5(8-4t)\right)dt=\int_{0}^{1}(43-18t)\,dt=43-9=34[/tex]

Solution:
Here [tex]P=2x-y[/tex] and [tex]Q=2x+y[/tex] are polynomials, hence continuously differentiable in the whole plane, and the direction [tex]O\to A\to B\to O[/tex] is counterclockwise (the triangle stays on the left). So Green's formula applies, where [tex]D[/tex] is the triangle:
[tex]\oint_L P\,dx+Q\,dy=\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\,dy=\iint_D\left(2-(-1)\right)dx\,dy=3\iint_D dx\,dy[/tex]
The side [tex]OB[/tex] lies on the line [tex]y=x[/tex], so [tex]D=\{0\le x\le 2,\ 0\le y\le x\}[/tex] is a right triangle with legs [tex]OA=AB=2[/tex] and area [tex]\frac{2\times 2}{2}=2[/tex]. Hence
[tex]\oint_L (2x-y)\,dx+(2x+y)\,dy=3\times 2=6[/tex]
Check by direct computation: on [tex]OA[/tex] ([tex]y=0,\ dy=0[/tex]) we get [tex]\int_0^2 2x\,dx=4[/tex]; on [tex]AB[/tex] ([tex]x=2,\ dx=0[/tex]) we get [tex]\int_0^2(4+y)\,dy=10[/tex]; on [tex]BO[/tex] ([tex]y=x[/tex], [tex]x[/tex] from [tex]2[/tex] to [tex]0[/tex]) we get [tex]\int_2^0 4x\,dx=-8[/tex]; in total [tex]4+10-8=6[/tex].

Solution:
The surface [tex]S[/tex] is not closed, so the Gauss-Ostrogradsky formula cannot be applied to it directly, but the flux [tex]\Phi[/tex] is easy to compute directly. On [tex]S[/tex] the unit normal is [tex]\vec{n}=(0, 0, 1)[/tex] and [tex]dS=dx\,dy[/tex], so only the third component of the field, taken at [tex]z=0[/tex], matters:
[tex]\Phi=\iint_{x^2+y^2\le 1}\left(z^2+x^2\right)\Big|_{z=0}dx\,dy=\iint_{x^2+y^2\le 1}x^2\,dx\,dy[/tex]
In polar coordinates [tex]x=r\cos\varphi,\ y=r\sin \varphi[/tex] (Jacobian [tex]r[/tex]):
[tex]\Phi=\int_0^{2\pi}\cos^2\varphi\,d\varphi\int_0^1 r^3\,dr=\pi\times \frac{1}{4}=\frac{\pi}{4}[/tex]

Solution:
Parametrize the circle counterclockwise:
[tex]x=\cos t,\quad y=\sin t,\quad 0\le t\le 2\pi[/tex]
Then [tex]dx=-\sin t\,dt[/tex], [tex]dy=\cos t\,dt[/tex], [tex]x^2+y^2=1[/tex] and
[tex]x\,dy-y\,dx=\left(\cos^2t+\sin ^2t\right)dt=dt\quad\Rightarrow\quad\oint_L \frac{x\,dy-y\,dx}{x^2+y^2}=\int_{0}^{2\pi}dt=2\pi[/tex]
Remark: for [tex]P=-\frac{y}{x^2+y^2}[/tex] and [tex]Q=\frac{x}{x^2+y^2}[/tex] we have [tex]\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}=\frac{y^2-x^2}{(x^2+y^2)^2}[/tex] for [tex](x,y)\ne(0,0)[/tex], but Green's formula cannot be applied to the unit disk, because [tex]P[/tex] and [tex]Q[/tex] are not defined at the origin, which lies inside [tex]L[/tex]. That is why the result is not [tex]0[/tex].