Solution. Completing the square in each variable, \[4(x-2)^2 - 16 + (y-3)^2 - 9 - 4(z+2)^2 + 16 + 9 = 0 ,\] \[4(x-2)^2 + (y-3)^2 - 4(z+2)^2 = 0 .\] The right-hand side is zero, so this is a cone, not a hyperboloid, with vertex \((2,3,-2)\) and axis parallel to the \(z-\)axis.

This is the second question in the course turning on that distinction, and the lesson is the same: check the constant before naming the surface. A positive constant would give a hyperboloid of one sheet, a negative one two sheets.

Solution. The inner integral cannot be done as written, since \(\sec ^2(y^2)\) has no elementary antiderivative in \(y\). Reverse the order. The region \[0\leq x\leq \frac {1}{\sqrt 2},\qquad x^2\leq y\leq \frac 12\] is the same as \[0\leq y\leq \frac 12,\qquad 0\leq x\leq \sqrt y .\] Then \[\int _0^{1/2}\int _0^{\sqrt y}x\sec ^2\big (y^2\big )dx\,dy = \int _0^{1/2}\frac {y}{2}\sec ^2\big (y^2\big )dy = \frac 14\Big [\tan \big (y^2\big )\Big ]_0^{1/2} = \frac {\tan \frac 14}{4}\approx 0.0638 .\] The factor of \(y\) produced by the inner integration is exactly what the substitution \(w = y^2\) needs.

Solution. With \(z_x = 2x\) and \(z_y = 2y\), \[dS = \sqrt {1+4x^2+4y^2}\;dA = \sqrt {1+4r^2}\;r\,dr\,d\theta .\] The planes \(z=1\) and \(z=2\) give \(r=1\) and \(r=\sqrt 2\), and the first octant restricts \(\theta \) to \(\Big [0,\dfrac {\pi }{2}\Big ]\). Hence \[S = \int _0^{\pi /2}\int _1^{\sqrt 2}\sqrt {1+4r^2}\;r\,dr\,d\theta = \frac {\pi }{2}\cdot \frac {1}{12}\Big [\big (1+4r^2\big )^{3/2}\Big ]_1^{\sqrt 2} = \frac {\pi \big (27-5\sqrt 5\big )}{24}\approx 2.07 .\]

Solution.

(a)

The four boundary curves become \(u=3\), \(u=5\), \(v=1\) and \(v=2\), so \(R'\) is the rectangle \([3,5]\times [1,2]\) — which is the whole point of the substitution.

(b)

\[\frac {\partial (u,v)}{\partial (x,y)} = \begin {vmatrix} y & x\\ -\dfrac {y}{x^2} & \dfrac 1x\end {vmatrix} = \frac {y}{x} + \frac {y}{x} = \frac {2y}{x} = 2v .\] The Jacobian needed for the change of variables is the reciprocal, \[\frac {\partial (x,y)}{\partial (u,v)} = \frac {1}{2v},\] which is the one that multiplies \(du\,dv\).

(c)

The integrand is \(v^2\sin v^2\), so \[\iint _R v^2\sin \big (v^2\big )dA = \int _3^5\int _1^2 v^2\sin \big (v^2\big )\frac {1}{2v}\,dv\,du = \int _3^5\int _1^2\frac {v\sin \big (v^2\big )}{2}dv\,du .\] The \(u\) integral contributes a factor of \(2\), and substituting \(w = v^2\), \[= 2\cdot \frac 14\Big [-\cos \big (v^2\big )\Big ]_1^2 = \frac {\cos 1 - \cos 4}{2}\approx 0.597 .\] Note the \(\dfrac {1}{2v}\) from the Jacobian is exactly what turns \(v^2\sin v^2\) into something integrable.

Solution. In cylindrical coordinates the cone is \(z=r\) and the cylinder \(r=a\), and the first octant gives \(0\leq \theta \leq \dfrac {\pi }{2}\). Hence \[V = \int _0^{\pi /2}\int _0^a\int _0^r dz\;r\,dr\,d\theta = \frac {\pi }{2}\int _0^a r^2dr = \frac {\pi a^3}{6}.\]