Solution

$\iint_R y\sin(ny)\,dA=\int_{1}^{2}\!\!dx\int_{0}^{\pi} y\sin(ny)\,dy$

$=\int_{0}^{\pi} y\sin(ny)\,dy$

Let $u=y,\; dv=\sin(ny)\,dy \Rightarrow v=-\frac{\cos(ny)}{n}$

$\int_{0}^{\pi} y\sin(ny)\,dy=\left[-\frac{y\cos(ny)}{n}\right]_{0}^{\pi}+\frac{1}{n}\int_{0}^{\pi}\cos(ny)\,dy$

$=-\frac{\pi\cos(n\pi)}{n}+\frac{1}{n^2}\left[\sin(ny)\right]_{0}^{\pi}=-\frac{\pi(-1)^n}{n}$

Solution

First octant: $0\le x\le 2$, $0\le y\le 3$, $0\le z\le 9-y^2$

$V=\int_{0}^{2}\!\!dx\int_{0}^{3}(9-y^2)\,dy=2\left[9y-\frac{y^3}{3}\right]_{0}^{3}=2(27-9)=36$

Solution

Use polar: $x=r\cos\theta,\; y=r\sin\theta$

Cylinder: $r^2=2r\cos\theta \Rightarrow r=2\cos\theta,\; -\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$

$V=\iint_D (x^2+y^2)\,dA=\int_{-\pi/2}^{\pi/2}\int_{0}^{2\cos\theta} r^2\cdot r\,dr\,d\theta$

$=\int_{-\pi/2}^{\pi/2}\left[\frac{r^4}{4}\right]_{0}^{2\cos\theta}d\theta=\int_{-\pi/2}^{\pi/2}4\cos^4\theta\,d\theta$

$=8\int_{0}^{\pi/2}\cos^4\theta\,d\theta=8\cdot\frac{3\pi}{16}=\frac{3\pi}{2}$

Solution

Using Green’s theorem:

$\oint_C (M\,dx+N\,dy)=\iint_D\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dA$

Here $M=y^2,\; N=x^2$ so $\frac{\partial N}{\partial x}=2x,\; \frac{\partial M}{\partial y}=2y$

Integral $=\iint_D 2(x-y)\,dA$ over triangle with vertices $(0,0),(1,0),(0,1)$

By symmetry $\iint_D x\,dA=\iint_D y\,dA$ so $\iint_D (x-y)\,dA=0$

Hence value $=0$

Solution

Solution

Solution

Solution

Solution

Actual integral:

$\int_{0}^{1} (x^3 - cx^2)\,dx = \left[\frac{x^4}{4} - \frac{cx^3}{3}\right]_0^1 = \frac{1}{4} - \frac{c}{3}$

Trapezoidal rule with one interval:

$T = \frac{1}{2}[f(0) + f(1)]$

$f(0) = 0$

$f(1) = 1 - c$

$T = \frac{1}{2}(1 - c)$

Since exact:

$\frac{1}{4} - \frac{c}{3} = \frac{1}{2}(1 - c)$

Solving:

$\frac{1}{4} - \frac{c}{3} = \frac{1}{2} - \frac{c}{2}$

Multiply by 12:

$3 - 4c = 6 - 6c$

$2c = 3$

$c = 1.5$