« Tricks of Integration (1) – The Koop Subs
Mid-Term 2 Solution »
h1

Math 324 Mid-Term 1 Solution

四月 18, 2009

1) (a) V(S) = \int_{-1}^{1}\int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}} \int_{x^2 + y^2}^{\sqrt{x^2 + y^2}}1 \, dz \, dy \, dx.

(b and c) \begin{aligned} V(S) &= \int_{0}^{2\pi}\int_{0}^{1}\int_{r^2}^{r}r \, dz \, dr \, d\theta\\ &= 2\pi \int_{0}^{1} r^2 - r^3 dr \\ &= 2\pi (\frac{1}{3} -\frac{1}{4}) = \frac{\pi}{6}.\end{aligned}

2) (a) The triple integral is maximized when the integrand is everywhere non-negative. i.e. when 1 - x^2 - y^2/4 - z^2/9 \ge 0. Equivalently, x^2 + y^2/4 + z^2/9 \le 1, which is an ellipsoid.

(b) We let u = x, v = y/2, w = z/3. This maps the ellipsoid R to the unit sphere S.
The Jacobian is: \frac{\partial(u, v, w)}{\partial(x, y, z)} = \det \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} = 6.

By the change of variables formula, we have \iiint_{R} 1 dV = \iiint_{S} 6 dV_{S} = 6\frac{4 \pi}{3} = 8\pi.

(c) By the change of variables formula, \iiint_{R} 1 - x^2 - y^2/4 - z^2/9 dV = \iiint_{S} (1 - u^2 - v^2 - w^2)6 dV_{S}.

Switching to spherical yields: \begin{aligned} 6\int_{0}^{2 \pi}\int_{0}^{\pi} \int_{0}^{1} (1 - \rho^2)\rho^2\sin(\phi)\, d\rho \, d\phi \, d\theta &= 12\pi \int_{0}^{\pi} \int_{0}^{1} (\rho^2 - \rho^4)\sin(\phi) \, d\rho \, d\phi \\&= 12\pi (\frac{1}{3} - \frac{1}{5})\int_{0}^{\pi} \sin(\phi) \, d\phi \\&= \frac{16 \pi}{5}. \end{aligned}

3) (a) Switching order of integration yields: I = \int_{-\pi/2}^{\pi/2} \int_{0}^{\cos{x}} \frac{2y}{1 + 2^x} \, dy \, dx = \int_{-\pi/2}^{\pi/2} \frac{\cos^2{x}}{1 + 2^x} dx.

Let u = -x, we have I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2{u}}{1 + 2^{-u}}du = \int_{-\pi/2}^{\pi/2} \frac{2^u\cos^2{u}}{1 + 2^u} du = \int_{-\pi/2}^{\pi/2} \frac{2^x \cos^{2}{x}}{1 + 2^x} dx.

Therefore, 2I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2{x}}{1 + 2^x} + \frac{2^x \cos^2{x}}{1 + 2^x} dx = \int_{-\pi/2}^{\pi/2} \cos^2{x} dx = 2\int_{0}^{\pi/2} \cos^2{x} dx.

Finally, I = \int_{0}^{\pi/2} \cos^2{x}dx = \frac{\pi}{4} by using the koop sub u = \pi/2 - x.

3) (b) Notice that x^2 + (y - x)^2 + y^2 = 2(x^2 - xy + y^2) = 2[(x - y/2)^2 + 3y^2/4] = u^2 + v^2, where u = \sqrt{2}(x - y/2), v = \sqrt{6}y/2.

The Jacobian is \frac{\partial(u, v)}{\partial(x, y)} = \det \begin{pmatrix} \sqrt{2} & -\sqrt{2}/2 \\ 0 & \sqrt{6}/2 \end{pmatrix} = \sqrt{3}. Therefore, \frac{\partial(x, y)}{\partial(u, v)} = \frac{1}{\sqrt{3}}.

By the change of variables formula, we have I = \frac{1}{\sqrt{3}}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(u^2 + v^2)} du dv = \frac{1}{\sqrt{3}} \left(\int_{-\infty}^{\infty} e^{-u^2} du\right)^2 = \frac{1}{\sqrt{3}} \pi since \int_{-\infty}^{\infty} e^{-u^2} du = \sqrt{\pi} as shown in class. (or we can easily compute it by switching to polar coordinates.)

Posted in Calculus, Teaching |

Leave a Comment