Solution.

(a)

\(2x^2 = 9-x^2\) gives \(3x^2 = 9\), so \(x = \pm \sqrt 3\) and \(y = 6\). In the first quadrant the meeting point is \(\big (\sqrt 3, 6\big )\).

(b)

\(R\) is bounded below by the rising parabola \(y=2x^2\), above by the falling parabola \(y = 9-x^2\), and on the left by the \(y-\)axis, from \(x=0\) to \(x=\sqrt 3\).

(c)

\[A = \int _0^{\sqrt 3}\int _{2x^2}^{9-x^2}dy\,dx .\]

(d)

Integrating \(x\) first requires splitting at \(y=6\), because below that height the right boundary is the lower parabola and above it the upper one: \[A = \int _0^6\int _0^{\sqrt {y/2}}dx\,dy + \int _6^9\int _0^{\sqrt {9-y}}dx\,dy .\] That the second form needs two integrals and the first only one is the whole argument for choosing the order carefully.

(e)

\[A = \int _0^{\sqrt 3}\big (9-3x^2\big )dx = \Big [9x - x^3\Big ]_0^{\sqrt 3} = 9\sqrt 3 - 3\sqrt 3 = 6\sqrt 3 .\]

(f)

\(R'\) is symmetric about the \(y-\)axis and consists of two copies of \(R\), so its area is \(12\sqrt 3\).

(g)

This is the hint. The integrand \(xy^2\) is odd in \(x\), while \(R'\) is symmetric about the \(y-\)axis, so the contributions from \(x>0\) and \(x<0\) cancel exactly: \[\iint _{R'}xy^2\,dA = 0 ,\] with no integration required.

Solution.

(a)

\(\dfrac {d}{dx}\sin x^2 = 2x\cos x^2\) by the chain rule.

(b)

Not in elementary terms. The antiderivative of \(\cos x^2\) is the Fresnel integral, which is not expressible with the standard functions. Part (a) shows that \(x\cos x^2\) can be integrated — the extra factor of \(x\) is exactly what the chain rule supplies — and that is the hint for (c).

(c)

As written, the inner integral is \(\int \cos x^2dx\), which cannot be done. So reverse the order. The region is \[0\leq y\leq 2,\qquad y^2\leq x\leq 4 ,\] which is the same as \[0\leq x\leq 4,\qquad 0\leq y\leq \sqrt x .\] Then \[\int _0^4\int _0^{\sqrt x}y\cos x^2\,dy\,dx = \int _0^4\frac {x}{2}\cos x^2\,dx = \Big [\frac {\sin x^2}{4}\Big ]_0^4 = \frac {\sin 16}{4}\approx -0.072 .\] The factor of \(x\) produced by the inner integration is what makes the outer one possible.

Solution. Again the inner integral is impossible as written, so reverse the order. The region \(0\leq x\leq 1\), \(x^3\leq y\leq 1\) is the same as \(0\leq y\leq 1\), \(0\leq x\leq y^{1/3}\). Hence \[\int _0^1\int _0^{y^{1/3}}x^5\sin \big (y^3\big )dx\,dy = \int _0^1\frac {y^2}{6}\sin \big (y^3\big )dy .\] Now substitute \(w = y^3\), \(dw = 3y^2dy\): \[= \frac {1}{18}\int _0^1\sin w\,dw = \frac {1-\cos 1}{18}\approx 0.0255 .\]

Solution. The slanted edge joins \((4,1)\) to \((1,2)\), with slope \(\dfrac {2-1}{1-4} = -\dfrac 13\), so its equation is \[y = 1 - \frac {x-4}{3} = \frac {7-x}{3},\] which gives \(y=2\) at \(x=1\) and \(y=1\) at \(x=4\), as required. Hence \[V = \int _1^4\int _1^{(7-x)/3}xy\,dy\,dx = \int _1^4\frac {x}{2}\Big [\Big (\frac {7-x}{3}\Big )^2 - 1\Big ]dx .\] Expanding, \(\Big (\dfrac {7-x}{3}\Big )^2 - 1 = \dfrac {49-14x+x^2-9}{9} = \dfrac {x^2-14x+40}{9}\), so \[V = \frac {1}{18}\int _1^4\big (x^3-14x^2+40x\big )dx = \frac {1}{18}\Big [\frac {x^4}{4}-\frac {14x^3}{3}+20x^2\Big ]_1^4 = \frac {31}{8}.\]

Solution.

(a)

The region lies between the lines \(y=x\) and \(y=2x\) for \(0\leq x\leq 1\), so it is the triangle with vertices \((0,0)\), \((1,1)\), \((1,2)\). Reading it in the other order, \(y\) runs from \(0\) to \(2\), and for a given \(y\) the variable \(x\) runs from \(\dfrac {y}{2}\) up to \(\min (y,1)\). The change happens at \(y=1\), so \[I = \int _0^1\int _{y/2}^{y}f\,dx\,dy + \int _1^2\int _{y/2}^{1}f\,dx\,dy .\]

(b)

With \(f\) independent of \(y\) the original order is easier: \[I = \int _0^1\big (2x-x\big )\sqrt {1-x^2}\,dx = \int _0^1x\sqrt {1-x^2}\,dx .\] Substituting \(w = 1-x^2\), \(dw = -2x\,dx\), \[I = \frac 12\int _0^1\sqrt w\,dw = \frac 13 .\]

Solution. Throughout, \(dA = r\,dr\,d\theta \) and the cone is \(z = r\).

(a)

\[V = \int _0^{2\pi }\int _0^2 r\cdot r\,dr\,d\theta = 2\pi \Big [\frac {r^3}{3}\Big ]_0^2 = \frac {16\pi }{3}.\]

(b)

The cone meets the sphere where \(r^2+r^2 = 1\), that is \(r = \dfrac {1}{\sqrt 2}\). Between the two surfaces the height is \(\sqrt {1-r^2}-r\), so \[V = \int _0^{2\pi }\int _0^{1/\sqrt 2}\Big (\sqrt {1-r^2}-r\Big )r\,dr\,d\theta = \frac {\pi \big (2-\sqrt 2\big )}{3}\approx 0.613 .\]

(c)

\[V = \int _0^{2\pi }\int _0^2\sqrt {16-r^2}\;r\,dr\,d\theta = 2\pi \Big [-\frac {\big (16-r^2\big )^{3/2}}{3}\Big ]_0^2 = \frac {2\pi }{3}\Big (64-24\sqrt 3\Big ) = \frac {16\pi \big (8-3\sqrt 3\big )}{3} \approx 47.0 .\]

Solution. Take the lamina as the upper half of the disc of radius \(a\), so the density is \(\delta = kr\) in polar coordinates. By symmetry \(\bar x = 0\).

\[m = \int _0^{\pi }\int _0^a kr\cdot r\,dr\,d\theta = k\pi \frac {a^3}{3}.\] The moment about the \(x-\)axis is \[M_x = \int _0^{\pi }\int _0^a (r\sin \theta )\,kr\cdot r\,dr\,d\theta = k\frac {a^4}{4}\int _0^{\pi }\sin \theta \,d\theta = \frac {ka^4}{2}.\] Hence \[\bar y = \frac {M_x}{m} = \frac {ka^4/2}{k\pi a^3/3} = \frac {3a}{2\pi } \approx 0.477a ,\] and the centre of mass is \(\Big (0, \dfrac {3a}{2\pi }\Big )\). For comparison, a lamina of uniform density has \(\bar y = \dfrac {4a}{3\pi }\approx 0.424a\): making the density grow with \(r\) pushes the centre of mass outward, as it should.

Solution. The \(x\) limits give \(x^2 = 4y-y^2\), that is \[x^2+y^2 = 4y ,\] which in polar form is \(r = 4\sin \theta \) — but note this is the circle \(x^2+(y-2)^2 = 4\), of radius \(2\) centred at \((0,2)\).

The \(y\) limits restrict to \(2\leq y\leq 4\), the upper half of that disc. In polar coordinates \(y\geq 2\) reads \(r\sin \theta \geq 2\), that is \(r\geq 2\csc \theta \). So \[2\csc \theta \leq r \leq 4\sin \theta ,\] and such \(r\) exist only when \(2\csc \theta \leq 4\sin \theta \), that is \(\sin ^2\theta \geq \dfrac 12\), giving \(\dfrac {\pi }{4}\leq \theta \leq \dfrac {3\pi }{4}\). Hence \[\int _2^4\int _{-\sqrt {4y-y^2}}^{\sqrt {4y-y^2}}ye^x\,dx\,dy = \int _{\pi /4}^{3\pi /4}\int _{2\csc \theta }^{4\sin \theta } \big (r\sin \theta \big )e^{r\cos \theta }\,r\,dr\,d\theta .\]

Solution.

(a)

The cone makes an angle \(\dfrac {\pi }{4}\) with the positive \(z-\)axis, so \[R = \Big \{(\rho ,\phi ,\theta ): 0\leq \rho \leq 3,\ 0\leq \phi \leq \frac {\pi }{4},\ 0\leq \theta \leq 2\pi \Big \},\] the ice-cream-cone region inside the sphere and above the cone.

(b)

\[V = \int _0^{2\pi }\int _0^{\pi /4}\int _0^3\rho ^2\sin \phi \,d\rho \,d\phi \,d\theta = 2\pi \cdot 9\cdot \Big [-\cos \phi \Big ]_0^{\pi /4} = 18\pi \Big (1-\frac {\sqrt 2}{2}\Big )\approx 16.6 .\]

(c)

The distance from the \(z-\)axis is \(\rho \sin \phi \), so the density is \(\dfrac {1}{\rho \sin \phi }\) and \[m = \int _0^{2\pi }\int _0^{\pi /4}\int _0^3 \frac {\rho ^2\sin \phi }{\rho \sin \phi }\,d\rho \,d\phi \,d\theta = \int _0^{2\pi }\int _0^{\pi /4}\int _0^3\rho \,d\rho \,d\phi \,d\theta .\] The \(\sin \phi \) of the volume element cancels against the density, leaving \[m = 2\pi \cdot \frac {\pi }{4}\cdot \frac 92 = \frac {9\pi ^2}{4}\approx 22.2 .\]

Solution.

(a) and (b)

Completing the square, \(x^2+(y-1)^2 = 1\): a circle of radius \(1\) centred \((0,1)\), which passes through the origin and lies wholly inside the disc of radius \(2\). In polar coordinates \(r^2 = 2r\sin \theta \), so \[r = 2\sin \theta ,\qquad 0\leq \theta \leq \pi .\] \(R\) is the solid under the hemisphere of radius \(2\) and above that smaller circular region.

(c)

The hemisphere has volume \(\dfrac 23\pi (2)^3 = \dfrac {16\pi }{3}\), so half of it is \(\dfrac {8\pi }{3}\approx 8.38\).

(d)

\[V = \int _0^{\pi }\int _0^{2\sin \theta }\sqrt {4-r^2}\;r\,dr\,d\theta = \int _0^{\pi }\Big [-\frac {\big (4-r^2\big )^{3/2}}{3}\Big ]_0^{2\sin \theta }d\theta = \frac {8}{3}\int _0^{\pi }\Big (1-\left |\cos \theta \right |^3\Big )d\theta .\] Evaluating, \(\displaystyle \int _0^{\pi }\left |\cos \theta \right |^3d\theta = \frac 43\), so \[V = \frac 83\Big (\pi - \frac 43\Big ) = \frac {8\pi }{3} - \frac {32}{9} \approx 4.82 .\]

Comparison

This is not half the hemisphere. The two would agree only if the circle \(r = 2\sin \theta \) cut the disc of radius \(2\) in half, and it does not: it is a circle of radius \(1\) sitting inside the disc of radius \(2\), touching it at one point. Its area is \(\pi \) against the disc’s \(4\pi \), a quarter, not a half.

(e)

The correct volume is therefore \(\dfrac {8\pi }{3}-\dfrac {32}{9}\approx 4.82\), as computed in (d).

Solution.

(a)

Substituting \(x^2+y^2 = 2z\) into the sphere gives \(2z+z^2 = 8\), so \(z^2+2z-8 = 0\) and \((z+4)(z-2) = 0\). Only \(z=2\) is admissible, and then \(x^2+y^2 = 4\): the curve of intersection is the circle of radius \(2\) at height \(2\).

(b)

In spherical coordinates the density is \(e^{\rho ^3}\), and the first octant is \(0\leq \phi \leq \dfrac {\pi }{2}\), \(0\leq \theta \leq \dfrac {\pi }{2}\), with \(\rho \) up to \(\sqrt 8 = 2\sqrt 2\). So \[m = \int _0^{\pi /2}\int _0^{\pi /2}\int _0^{2\sqrt 2} e^{\rho ^3}\rho ^2\sin \phi \,d\rho \,d\phi \,d\theta .\] The angular integrals give \(\dfrac {\pi }{2}\) and \(1\), and the substitution \(w = \rho ^3\), \(dw = 3\rho ^2d\rho \) handles the rest: \[m = \frac {\pi }{2}\cdot \frac 13\Big [e^{\rho ^3}\Big ]_0^{2\sqrt 2} = \frac {\pi }{6}\Big (e^{16\sqrt 2}-1\Big ).\] The exponent \(\rho ^3\) matching the \(\rho ^2\) of the volume element is the whole reason this is integrable.

(c)

Cylindrical coordinates suit the paraboloid. Within it and in the first octant, \(\theta \) runs from \(0\) to \(\dfrac {\pi }{2}\), \(r\) from \(0\) to \(2\), and \(z\) from the paraboloid \(\dfrac {r^2}{2}\) up to the sphere \(\sqrt {8-r^2}\): \[V = \int _0^{\pi /2}\int _0^2\int _{r^2/2}^{\sqrt {8-r^2}}r\,dz\,dr\,d\theta = \frac {\pi }{2}\int _0^2\Big (\sqrt {8-r^2}-\frac {r^2}{2}\Big )r\,dr .\] The first part gives \(\dfrac 13\big (8\sqrt 8 - 4^{3/2}\big )\) and the second \(2\), so \[V = \frac {\pi }{2}\Big (\frac {16\sqrt 2-8}{3} - 2\Big ) = \frac {\pi }{6}\Big (16\sqrt 2 - 14\Big )\approx 4.52 .\]

Solution.

(a)

\[\overline {r}_u = \langle 2u, v, 0\rangle ,\qquad \overline {r}_v = \langle 0, u, v\rangle ,\] \[\overline {r}_u\times \overline {r}_v = \langle v^2,\ -2uv,\ 2u^2\rangle ,\] whose length is \[\sqrt {v^4+4u^2v^2+4u^4} = \sqrt {\big (v^2+2u^2\big )^2} = v^2+2u^2 ,\] a perfect square, which is what makes the integral elementary. Hence \[S = \int _0^2\int _0^1\big (v^2+2u^2\big )du\,dv = \int _0^2\Big (v^2+\frac 23\Big )dv = \frac 83+\frac 43 = 4 .\]

(b)

With \(z = xy\), \(z_x = y\) and \(z_y = x\), so \[S = \iint _D\sqrt {1+x^2+y^2}\,dA = \int _0^{2\pi }\int _0^1\sqrt {1+r^2}\;r\,dr\,d\theta = 2\pi \Big [\frac {\big (1+r^2\big )^{3/2}}{3}\Big ]_0^1 = \frac {2\pi \big (2\sqrt 2-1\big )}{3}\approx 3.83 .\]

Solution. The region sits over the rectangle \(0\leq x\leq 2\), \(0\leq y\leq 6\) — the upper limit on \(x\) being where \(z=4-x^2\) meets \(z=0\) — with height \(4-x^2\).

\[V = \int _0^6\int _0^2\big (4-x^2\big )dx\,dy = 6\Big [4x-\frac {x^3}{3}\Big ]_0^2 = 6\cdot \frac {16}{3} = 32 .\] For the centroid, each coordinate is the corresponding average: \[\bar x = \frac {1}{V}\int _0^6\int _0^2x\big (4-x^2\big )dx\,dy = \frac {6\cdot 4}{32} = \frac 34 ,\] \[\bar y = 3 \quad \text {by symmetry in } y ,\] \[\bar z = \frac {1}{V}\int _0^6\int _0^2\frac {\big (4-x^2\big )^2}{2}dx\,dy = \frac {8}{5}.\] So the centroid is \(\Big (\dfrac 34,\ 3,\ \dfrac 85\Big )\).

Solution. The paraboloid opens along the \(x-\)axis. Using polar coordinates in the \(yz-\)plane, with \(y = r\cos \theta \) and \(z = r\sin \theta \), the paraboloid is \(x = 4r^2\) and it meets \(x=4\) where \(r=1\). Hence \[\iiint _E x\,dV = \int _0^{2\pi }\int _0^1\int _{4r^2}^{4}x\,dx\;r\,dr\,d\theta = \int _0^{2\pi }\int _0^1\frac {16-16r^4}{2}\,r\,dr\,d\theta \] \[= 8\cdot 2\pi \int _0^1\big (r-r^5\big )dr = 16\pi \Big (\frac 12-\frac 16\Big ) = \frac {16\pi }{3}\approx 16.8 .\]

Solution. dividing by \(25\), the ellipse is \(x^2+\dfrac {y^2}{25} = 1\), so the substitution \[x = u,\qquad y = 5v\] turns it into the unit circle \(u^2+v^2 = 1\). The Jacobian is \[\frac {\partial (x,y)}{\partial (u,v)} = \begin {vmatrix}1 & 0\\ 0 & 5\end {vmatrix} = 5 ,\] so \(dA = 5\,du\,dv\) and \[\iint _R y^2\,dA = \iint _{u^2+v^2\leq 1}25v^2\cdot 5\,du\,dv = 125\iint _{u^2+v^2\leq 1}v^2\,du\,dv .\] In polar coordinates the remaining integral is \[\int _0^{2\pi }\int _0^1 r^2\sin ^2\theta \;r\,dr\,d\theta = \frac 14\int _0^{2\pi }\sin ^2\theta \,d\theta = \frac {\pi }{4},\] so the answer is \[125\cdot \frac {\pi }{4} = \frac {125\pi }{4}\approx 98.2 .\]