Multiple, Line and Surface Integrals: Difficult Problems with Solutions

Solution:
Completing the square,
[tex]x^2+y^2\le x\iff \left(x-\frac{1}{2}\right)^2+y^2\le\frac{1}{4}[/tex]
so [tex]D[/tex] is the disk with center [tex]\left(\frac{1}{2}, 0\right)[/tex] and radius [tex]\frac{1}{2}[/tex], whose boundary passes through the origin. We use polar coordinates with pole at the origin: [tex]x=r\cos\varphi,\ y=r\sin \varphi[/tex], Jacobian [tex]r[/tex], and [tex]\sqrt{x^2+y^2}=r[/tex]. The condition becomes [tex]r^2\le r\cos\varphi[/tex], i.e. [tex]0\le r\le\cos\varphi[/tex], which requires [tex]\cos\varphi\ge 0[/tex], i.e. [tex]-\frac{\pi}{2}\le\varphi\le\frac{\pi}{2}[/tex]. Hence the integral [tex]I[/tex] equals
[tex]I=\int_{-\pi/2}^{\pi/2}\int_{0}^{\cos\varphi}r^2\,dr\,d\varphi=\frac{1}{3}\int_{-\pi/2}^{\pi/2}\cos^3\varphi\,d\varphi=\frac{1}{3}\left[\sin \varphi-\frac{\sin ^3\varphi}{3}\right]_{-\pi/2}^{\pi/2}=\frac{1}{3}\times \frac{4}{3}=\frac{4}{9}[/tex]
Note: the bound for [tex]r[/tex] must depend on [tex]\varphi[/tex], because the pole lies on the boundary circle, not at its center; taking [tex]0\le r\le 1[/tex] is a common mistake.

Solution:
The region of integration is
[tex]D=\left\{-2\le y\le 2,\ y^2-2y-3\le x\le 1-2y\right\}[/tex]
Its left boundary is the parabola [tex]x=(y-1)^2-4[/tex] with vertex [tex](-4, 1)[/tex], opening to the right; its right boundary is the line [tex]x=1-2y[/tex], i.e. [tex]y=\frac{1-x}{2}[/tex]. They intersect where
[tex]y^2-2y-3=1-2y\iff y^2=4\iff y=\pm 2[/tex]
i.e. at the points [tex](5, -2)[/tex] and [tex](-3, 2)[/tex]. These are exactly the given limits for [tex]y[/tex], and for [tex]|y|\lt 2[/tex] the parabola lies to the left of the line (e.g. for [tex]y=0[/tex]: [tex]-3\lt 1[/tex]). Solving the equation of the parabola for [tex]y[/tex] gives its lower and upper branches:
[tex]y=1-\sqrt{x+4},\qquad y=1+\sqrt{x+4}[/tex]
In [tex]D[/tex] the variable [tex]x[/tex] runs from [tex]-4[/tex] (the vertex) to [tex]5[/tex]. For [tex]-4\le x\le -3[/tex] a vertical line meets [tex]D[/tex] between the two branches of the parabola (the upper branch reaches the corner [tex](-3, 2)[/tex] at [tex]x=-3[/tex]). For [tex]-3\le x\le 5[/tex] the lower boundary is still [tex]y=1-\sqrt{x+4}[/tex], while the upper one is the line [tex]y=\frac{1-x}{2}[/tex]. Therefore
[tex]\int_{-2}^{2}dy\int_{y^2-2y-3}^{1-2y}f(x,y)\,dx=\int_{-4}^{-3}dx\int_{1-\sqrt{x+4}}^{1+\sqrt{x+4}}f(x,y)\,dy+\int_{-3}^{5}dx\int_{1-\sqrt{x+4}}^{\frac{1-x}{2}}f(x,y)\,dy[/tex]
Check with [tex]f=1[/tex] (both sides give the area of [tex]D[/tex]):
[tex]\int_{-2}^{2}\left(4-y^2\right)dy=\frac{32}{3},\qquad \frac{4}{3}+\frac{28}{3}=\frac{32}{3}[/tex]

Solution:
The bounded solid is the part of the upper nappe of the cone ([tex]z\ge 0[/tex]) below the plane:
[tex]V:\ \sqrt{x^2+y^2}\le z\le 1[/tex]
In cylindrical coordinates [tex]x=r\cos\varphi,\ y=r\sin \varphi,\ z=z[/tex] (Jacobian [tex]r[/tex]) the integrand becomes [tex]r[/tex] and
[tex]V:\ 0\le\varphi\le 2\pi,\ 0\le r\le 1,\ r\le z\le 1[/tex]
Hence (the integrand times the Jacobian gives [tex]r^2[/tex])
[tex]\iiint_V \sqrt{x^2+y^2}\,dx\,dy\,dz=\int_{0}^{2\pi}d\varphi\int_{0}^{1}dr\int_{r}^{1}r^2\,dz=2\pi\int_{0}^{1}r^2(1-r)\,dr=2\pi\left(\frac{1}{3}-\frac{1}{4}\right)=\frac{\pi}{6}[/tex]

Solution:
For a fixed point [tex](x, y)[/tex] the variable [tex]z[/tex] runs from [tex]0[/tex] to [tex]3\sqrt{1-\frac{x}{7}-\frac{y}{10}}[/tex]; this is defined exactly when [tex](x, y)[/tex] lies in the triangle [tex]T[/tex] (the projection of the solid onto the plane [tex]Oxy[/tex], obtained for [tex]z=0[/tex]):
[tex]T=\left\{x\ge0,\ y\ge0,\ \frac{x}{7}+\frac{y}{10}\le1\right\},\qquad V=\iint_T 3\sqrt{1-\frac{x}{7}-\frac{y}{10}}\,dx\,dy[/tex]
Substitute [tex]x=7u,\ y=10v[/tex]; the Jacobian is [tex]70[/tex] and [tex]T[/tex] becomes the triangle [tex]u\ge0,\ v\ge0,\ u+v\le1[/tex]:
[tex]V=210\int_0^1du\int_0^{1-u}\sqrt{1-u-v}\,dv[/tex]
The inner integral is
[tex]\int_0^{1-u}\sqrt{1-u-v}\,dv=-\frac23(1-u-v)^{3/2}\Big|_{v=0}^{v=1-u}=\frac23(1-u)^{3/2}[/tex]
hence
[tex]V=210\int_0^1\frac23(1-u)^{3/2}\,du=210\times \frac23\times \frac25=56[/tex]

Solution:
Use spherical coordinates
[tex]x=\rho\sin \theta\cos\varphi,\ y=\rho\sin \theta\sin \varphi,\ z=\rho\cos\theta,\qquad dx\,dy\,dz=\rho^2\sin \theta\,d\rho\,d\theta\,d\varphi[/tex]
The first octant gives [tex]0\le\varphi\le\frac{\pi}{2}[/tex]. The condition [tex]z\ge\sqrt{x^2+y^2}[/tex] becomes [tex]\rho\cos\theta\ge\rho\sin \theta[/tex], i.e. [tex]0\le\theta\le\frac{\pi}{4}[/tex], and the sphere gives [tex]0\le\rho\le\sqrt2[/tex]. Since [tex]z^2=\rho^2\cos^2\theta[/tex], the integral splits into a product:
[tex]\iiint_V z^2\,dx\,dy\,dz=\int_0^{\pi/2}d\varphi\int_0^{\pi/4}\cos^2\theta\sin \theta\,d\theta\int_0^{\sqrt2}\rho^4\,d\rho[/tex]
The three factors are
[tex]\frac{\pi}{2},\qquad \left[-\frac{\cos^3\theta}{3}\right]_0^{\pi/4}=\frac13\left(1-\frac{\sqrt2}{4}\right)=\frac{4-\sqrt2}{12},\qquad \frac{(\sqrt2)^5}{5}=\frac{4\sqrt2}{5}[/tex]
Therefore
[tex]\iiint_V z^2\,dx\,dy\,dz=\frac{\pi}{2}\times \frac{4-\sqrt2}{12}\times \frac{4\sqrt2}{5}=\frac{\pi\left(16\sqrt2-8\right)}{120}=\frac{\pi\left(2\sqrt2-1\right)}{15}[/tex]

Solution:
The ellipse is [tex]\frac{x^2}{4}+y^2=1[/tex], so its arc in the first quadrant can be parametrized as follows (an integral of the first kind does not depend on the direction of traversal):
[tex]x=2\cos t,\quad y=\sin t,\quad 0\le t\le\frac{\pi}{2}[/tex]
Then
[tex]dl=\sqrt{x'^2+y'^2}\,dt=\sqrt{4\sin ^2t+\cos^2t}\,dt=\sqrt{1+3\sin ^2t}\,dt[/tex]
[tex]\int_L xy\,dl=\int_{0}^{\pi/2}2\sin t\cos t\,\sqrt{1+3\sin ^2t}\,dt[/tex]
Substitute [tex]u=\sin ^2t[/tex], [tex]du=2\sin t\cos t\,dt[/tex]; then [tex]u[/tex] runs from [tex]0[/tex] to [tex]1[/tex]:
[tex]\int_L xy\,dl=\int_{0}^{1}\sqrt{1+3u}\,du=\frac{2}{9}\left[(1+3u)^{3/2}\right]_{0}^{1}=\frac{2}{9}(8-1)=\frac{14}{9}[/tex]

Solution:
The spiral starts at the origin and stays inside the unit circle while [tex]\rho=2\varphi\le1[/tex], i.e. for [tex]0\le\varphi\le\frac12[/tex]. For a curve given in polar coordinates as [tex]\rho=\rho(\varphi)[/tex] the arc element is
[tex]dl=\sqrt{\rho^2+\rho'^2}\,d\varphi=\sqrt{4\varphi^2+4}\,d\varphi=2\sqrt{1+\varphi^2}\,d\varphi[/tex]
Since [tex]\varphi\in\left[0, \frac12\right]\subset\left[0, \frac{\pi}{2}\right)[/tex], the points of [tex]L[/tex] lie in the first quadrant and [tex]\arctan\frac{y}{x}=\arctan(\tan \varphi)=\varphi[/tex] (the origin is a single point and does not affect the integral). Hence the integral equals
[tex]\int_0^{1/2}2\varphi\sqrt{1+\varphi^2}\,d\varphi=\frac23\left(1+\varphi^2\right)^{3/2}\Big|_0^{1/2}=\frac23\left(\frac{5\sqrt5}{8}-1\right)=\frac{5\sqrt5-8}{12}[/tex]

Solution:
For a line integral of the second kind along a parametrized curve
[tex]\int_L P\,dx+Q\,dy=\int_{0}^{2\pi}\left(P\,x'(t)+Q\,y'(t)\right)dt[/tex]
Here [tex]x'(t)=3(1-\cos t)[/tex], [tex]y'(t)=3\sin t[/tex] and [tex]6-y=3+3\cos t[/tex], so
[tex](6-y)\,x'=9(1+\cos t)(1-\cos t)=9\sin ^2t,\qquad x\,y'=9(t-\sin t)\sin t=9t\sin t-9\sin ^2t[/tex]
and the integrand simplifies to [tex]9t\sin t[/tex]. Integrating by parts, [tex]\int t\sin t\,dt=-t\cos t+\sin t[/tex], hence
[tex]\int_L (6-y)\,dx+x\,dy=9\left[-t\cos t+\sin t\right]_{0}^{2\pi}=9\times (-2\pi)=-18\pi[/tex]

Solution:
[tex]S[/tex] is the graph of [tex]z=\sqrt{x^2+y^2}[/tex] over the quarter disk
[tex]D:\ x^2+y^2\le 1,\ x\ge 0,\ y\ge 0[/tex]
Since [tex]z_x=\frac{x}{\sqrt{x^2+y^2}}[/tex] and [tex]z_y=\frac{y}{\sqrt{x^2+y^2}}[/tex], we have [tex]z_x^2+z_y^2=1[/tex] and (the vertex is a single point and does not affect the integral)
[tex]dS=\sqrt{1+z_x^2+z_y^2}\,dx\,dy=\sqrt{2}\,dx\,dy[/tex]
Hence, passing to polar coordinates,
[tex]\iint_S xy\,dS=\sqrt{2}\iint_D xy\,dx\,dy=\sqrt{2}\int_{0}^{\pi/2}\cos\varphi\sin \varphi\,d\varphi\int_{0}^{1}r^3\,dr=\sqrt{2}\times \frac{1}{2}\times \frac{1}{4}=\frac{\sqrt{2}}{8}[/tex]

Solution:
The area of a surface given parametrically over the parameter rectangle [tex]D:\ 0\le u\le 1,\ 0\le v\le 2[/tex] is
[tex]S=\iint_D\left|\vec{r}_u×\vec{r}_v\right|du\,dv[/tex]
Here the partial derivatives and their cross product are
[tex]\vec{r}_u=\left(0,\ v,\ u\right),\qquad \vec{r}_v=\left(2v,\ u,\ 0\right),\qquad \vec{r}_u×\vec{r}_v=\left(-u^2,\ 2uv,\ -2v^2\right)[/tex]
so the length of the normal vector is
[tex]\left|\vec{r}_u×\vec{r}_v\right|=\sqrt{u^4+4u^2v^2+4v^4}=\sqrt{\left(u^2+2v^2\right)^2}=u^2+2v^2[/tex]
The parametrization is one-to-one on [tex]D[/tex], because for [tex]u,v\ge 0[/tex] the parameters are recovered from the point: [tex]u=\sqrt{2z}[/tex], [tex]v=\sqrt{x}[/tex]. Hence
[tex]S=\int_0^1du\int_0^2\left(u^2+2v^2\right)dv=\int_0^1\left(2u^2+\frac{16}{3}\right)du=\frac{2}{3}+\frac{16}{3}=6[/tex]