Mass, Center of Mass & Surface Area

The curves meet at $x = 0$ and $x = 1$, with $x^2 \le y \le x$ between. No circles: Cartesian.

$M = \displaystyle\int_0^1\int_{x^2}^{x} x\,dy\,dx$.

Inner: $x(x - x^2) = x^2 - x^3$. Outer: $\displaystyle\int_0^1 (x^2 - x^3)\,dx = \tfrac{1}{3} - \tfrac{1}{4} = \tfrac{1}{12}$.

A quarter of a ring, and $\rho = r$: polar, with $1 \le r \le 2$, $0 \le \theta \le \tfrac{\pi}{2}$.

$M = \displaystyle\int_0^{\pi/2}\int_1^2 r \cdot r\,dr\,d\theta$.

Inner: $\displaystyle\int_1^2 r^2\,dr = \tfrac{7}{3}$. Outer: $\tfrac{\pi}{2} \cdot \tfrac{7}{3} = \tfrac{7\pi}{6}$.

Expand: $x^2 + y^2 = 4x$, so $r^2 = 4r\cos\theta$ and $r = 4\cos\theta$. Polar: $0 \le r \le 4\cos\theta$, $-\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}$.

$M = \displaystyle\int_{-\pi/2}^{\pi/2}\int_0^{4\cos\theta} r \cdot r\,dr\,d\theta$.

Inner: $\tfrac{64}{3}\cos^3\theta$. Outer: $\tfrac{64}{3}\displaystyle\int_{-\pi/2}^{\pi/2}\cos^3\theta\,d\theta = \tfrac{64}{3} \cdot \tfrac{4}{3} = \tfrac{256}{9}$.

The region is symmetric about the $y$-axis and $\rho$ is constant, so $\bar x = 0$.

Uniform, so use the area: $A = \displaystyle\int_{-2}^{2}(4 - x^2)\,dx = \tfrac{32}{3}$.

$\displaystyle\iint_D y\,dA = \int_{-2}^{2}\int_0^{4 - x^2} y\,dy\,dx = \int_{-2}^{2}\tfrac{1}{2}(4 - x^2)^2\,dx = \tfrac{256}{15}$.

$\bar y = \dfrac{256/15}{32/3} = \dfrac{8}{5}$.

The density $x$ breaks the symmetry, so compute both. The triangle is $0 \le x \le 2$, $0 \le y \le 2 - x$.

$M = \displaystyle\int_0^2\int_0^{2 - x} x\,dy\,dx = \int_0^2 x(2 - x)\,dx = \tfrac{4}{3}$.

$\displaystyle\iint_D x\rho\,dA = \int_0^2 x^2(2 - x)\,dx = \tfrac{4}{3}$ and $\displaystyle\iint_D y\rho\,dA = \int_0^2 \tfrac{x(2 - x)^2}{2}\,dx = \tfrac{2}{3}$.

$\bar x = \dfrac{4/3}{4/3} = 1$ and $\bar y = \dfrac{2/3}{4/3} = \dfrac{1}{2}$.

Region and density are symmetric about the $y$-axis: $\bar x = 0$. Polar: $0 \le r \le 3$, $0 \le \theta \le \pi$, $\rho = r$.

$M = \displaystyle\int_0^{\pi}\int_0^3 r \cdot r\,dr\,d\theta = \pi \cdot 9 = 9\pi$.

$\displaystyle\iint_D y\rho\,dA = \int_0^{\pi}\int_0^3 (r\sin\theta)\,r \cdot r\,dr\,d\theta = 2 \cdot \tfrac{81}{4} = \tfrac{81}{2}$.

$\bar y = \dfrac{81/2}{9\pi} = \dfrac{9}{2\pi}$.

$\displaystyle\iint_D 2\,dA - \iint_D 3y\,dA$.

$D$ is symmetric about the $x$-axis and $3y$ is odd in $y$: $\displaystyle\iint_D 3y\,dA = 0$.

$\displaystyle\iint_D 2\,dA = 2 \cdot \pi(2)^2 = 8\pi$.

Nothing is left.

$\displaystyle\iint_D xy^2\,dA + \iint_D \sqrt{9 - x^2 - y^2}\,dA$.

$D$ is symmetric about the $y$-axis and $xy^2$ is odd in $x$: that piece is $0$.

No constant piece.

$z = \sqrt{9 - x^2 - y^2}$ over $D$ is the top half of a ball of radius $3$: volume $\tfrac{2}{3}\pi(3)^3 = 18\pi$.

$\displaystyle\iint_T x^3\,dA + \iint_T 5\,dA$.

$T$ is symmetric about the $y$-axis and $x^3$ is odd: $\displaystyle\iint_T x^3\,dA = 0$.

The area of $T$ is $\tfrac{1}{2}(2)(2) = 2$, so $\displaystyle\iint_T 5\,dA = 10$.

Nothing is left.

$A = 2 \cdot 3 = 6$.

$\displaystyle\int_0^2\int_0^3 xy\,dy\,dx = \int_0^2 x\,dx \cdot \int_0^3 y\,dy = 2 \cdot \tfrac{9}{2} = 9$.

$\dfrac{9}{6} = \dfrac{3}{2}$.

$A = \pi(3)^2 = 9\pi$.

Polar: $\displaystyle\int_0^{2\pi}\int_0^3 r^2 \cdot r\,dr\,d\theta = 2\pi \cdot \tfrac{81}{4} = \tfrac{81\pi}{2}$.

$\dfrac{81\pi/2}{9\pi} = \dfrac{9}{2}$.

$A = \tfrac{1}{2}(2)(2) = 2$.

$\displaystyle\int_0^2\int_0^{2 - x}(x + y)\,dy\,dx = \int_0^2 \left(x(2 - x) + \tfrac{(2 - x)^2}{2}\right)\,dx = \tfrac{4}{3} + \tfrac{4}{3} = \tfrac{8}{3}$.

$\dfrac{8/3}{2} = \dfrac{4}{3}$.

$z = 3 - 2x - y$.

$f_x = -2$, $f_y = -1$: $\sqrt{1 + 4 + 1} = \sqrt{6}$.

$D$ is the disk $x^2 + y^2 \le 4$, with area $4\pi$.

The integrand is constant: $S = \sqrt{6} \cdot 4\pi = 4\sqrt{6}\,\pi$.

Already $z = f(x, y) = xy$.

$f_x = y$, $f_y = x$: $\sqrt{1 + x^2 + y^2} = \sqrt{1 + r^2}$.

$D$: $0 \le r \le \sqrt{3}$, $0 \le \theta \le 2\pi$.

$S = \displaystyle\int_0^{2\pi}\int_0^{\sqrt{3}}\sqrt{1 + r^2}\,r\,dr\,d\theta = 2\pi \cdot \tfrac{1}{3}\Big[(1 + r^2)^{3/2}\Big]_0^{\sqrt{3}} = 2\pi \cdot \tfrac{7}{3} = \tfrac{14\pi}{3}$.

$z = 4 - y^2$.

$f_x = 0$, $f_y = -2y$: $\sqrt{1 + 4y^2}$.

$D$: $0 \le y \le 1$, $0 \le x \le y$. Integrating $x$ first puts a $y$ beside the root.

$S = \displaystyle\int_0^1\int_0^y \sqrt{1 + 4y^2}\,dx\,dy = \int_0^1 y\sqrt{1 + 4y^2}\,dy = \tfrac{1}{12}\Big[(1 + 4y^2)^{3/2}\Big]_0^1 = \tfrac{5\sqrt{5} - 1}{12}$.