Practice Exam 4

Multiple Integrals

Questions10

Time90 min

Points100

SketchRequired for Q2, 3, 7, 10

CoversUnit 4

1Iterated integral8 pts

Evaluate \(\displaystyle\int_0^1\!\!\int_0^2 (x^2y + 2y)\,dy\,dx\).

Solution

\[ \int_0^2 (x^2 + 2)\,y\,dy = (x^2 + 2)\left[\frac{y^2}{2}\right]_0^2 = 2(x^2 + 2) \] \[ \int_0^1 (2x^2 + 4)\,dx = \frac23 + 4 = \frac{14}{3} \]

Inner integral first, treating \(x\) as a constant. The integrand factors as \((x^2 + 2)\cdot y\) over a rectangle, so it is also \(\int_0^1(x^2+2)\,dx\cdot\int_0^2 y\,dy = \tfrac73\cdot2 = \tfrac{14}{3}\). ✓

2General region10 pts

Evaluate \(\iint_D 2xy\,dA\) where \(D\) is bounded by \(y = x^2\) and \(y = x\).

Solution

Sketch: the curves meet at \(x = 0\) and \(x = 1\); on \((0,1)\) the line is above the parabola. Vertical strips from \(y = x^2\) to \(y = x\):

\[ \int_0^1\!\!\int_{x^2}^{x} 2xy\,dy\,dx = \int_0^1 x\Big[y^2\Big]_{x^2}^{x}dx = \int_0^1 (x^3 - x^5)\,dx = \frac14 - \frac16 = \frac{1}{12} \]

Horizontal strips would run from \(x = y\) to \(x = \sqrt y\) for \(0\le y\le1\) — try it; same answer.

3Reversing the order10 pts

Evaluate \(\displaystyle\int_0^2\!\!\int_{y/2}^{1} e^{x^2}\,dx\,dy\).

Solution

\(\int e^{x^2}dx\) cannot be done. Recover the region: \(0\le y\le2\), \(y/2\le x\le1\) — the triangle with vertices \((0,0)\), \((1,0)\), \((1,2)\), below the line \(y = 2x\). Vertical strips: \(0\le x\le1\), \(0\le y\le2x\).

\[ \int_0^1\!\!\int_0^{2x} e^{x^2}\,dy\,dx = \int_0^1 2x\,e^{x^2}\,dx = \Big[e^{x^2}\Big]_0^1 = e - 1\approx 1.718 \]

The inner integral in \(y\) manufactured the factor \(2x\) that \(u = x^2\) needs. This is always the shape of a reverse-the-order problem.

4Polar coordinates10 pts

(a) Evaluate \(\iint_D e^{-(x^2+y^2)}\,dA\) over the disk \(x^2+y^2\le4\). (b) Find the area enclosed by the cardioid \(r = 1 + \cos\theta\).

Solution

\[ \text{(a)}\quad \int_0^{2\pi}\!\!\int_0^2 e^{-r^2}\,r\,dr\,d\theta = 2\pi\left[-\tfrac12 e^{-r^2}\right]_0^2 = \pi\left(1 - e^{-4}\right)\approx 3.08 \]

Without the Jacobian \(r\), the inner integral would be the impossible \(\int e^{-r^2}dr\). With it, \(u = -r^2\) works instantly.

\[ \text{(b)}\quad A = \iint dA = \int_0^{2\pi}\!\!\int_0^{1+\cos\theta} r\,dr\,d\theta = \frac12\int_0^{2\pi}(1 + \cos\theta)^2\,d\theta = \frac12\int_0^{2\pi}\left(1 + 2\cos\theta + \frac{1 + \cos2\theta}{2}\right)d\theta = \frac12\left(2\pi + 0 + \pi\right) = \frac{3\pi}{2} \]

The Calculus II polar-area formula \(\tfrac12\int r^2\,d\theta\) is just this double integral with the inner \(r\)-integration already done.

5Mass & centre of mass10 pts

A lamina occupies the triangle with vertices \((0,0)\), \((1,0)\), \((0,1)\) and has density \(\rho(x,y) = x + y\). Find its mass and centre of mass.

Solution

Region: \(0\le x\le1\), \(0\le y\le1-x\).

\[ m = \int_0^1\!\!\int_0^{1-x}(x + y)\,dy\,dx = \int_0^1\left[x(1-x) + \frac{(1-x)^2}{2}\right]dx = \int_0^1\left(\frac12 - \frac{x^2}{2}\right)dx = \frac12 - \frac16 = \frac13 \] \[ M_y = \iint x(x+y)\,dA = \int_0^1\left[x^2(1-x) + \frac{x(1-x)^2}{2}\right]dx = \int_0^1\left(\frac{x}{2} - \frac{x^3}{2}\right)dx = \frac14 - \frac18 = \frac18 \]

\(\bar x = M_y/m = \tfrac{1/8}{1/3} = \tfrac38\). By the symmetry of both region and density under \(x\leftrightarrow y\), \(\bar y = \tfrac38\) too. Centre of mass \((\tfrac38, \tfrac38)\).

Sensible? A uniform triangle would balance at \((\tfrac13, \tfrac13)\). The density grows away from the origin, so the balance point should shift outward. \(\tfrac38 > \tfrac13\). ✓

6Surface area10 pts

(a) Find the area of the part of the plane \(z = 2x + 3y\) that lies over the disk \(x^2+y^2\le1\). (b) Find the area of the part of the cone \(z = \sqrt{x^2+y^2}\) below \(z = 2\).

Solution

\[ \text{(a)}\quad \sqrt{1 + f_x^2 + f_y^2} = \sqrt{1 + 4 + 9} = \sqrt{14},\qquad A = \sqrt{14}\cdot\pi(1)^2 = \sqrt{14}\,\pi\approx 11.8 \]

A tilted plane has a constant stretch factor, so the area is the factor times the shadow's area.

\[ \text{(b)}\quad f_x = \frac{x}{\sqrt{x^2+y^2}},\ f_y = \frac{y}{\sqrt{x^2+y^2}},\qquad \sqrt{1 + f_x^2 + f_y^2} = \sqrt{1 + \frac{x^2 + y^2}{x^2+y^2}} = \sqrt2 \] \[ A = \sqrt2\cdot\pi(2)^2 = 4\sqrt2\,\pi\approx 17.8 \]

Also constant — a cone is a rolled-up flat sector. Check with geometry: lateral area \(\pi r\ell\) with \(r = 2\), slant height \(\ell = 2\sqrt2\): \(4\sqrt2\pi\). ✓

7Triple integral, Cartesian10 pts

Let \(E\) be the tetrahedron bounded by the coordinate planes and \(x + y + z = 1\). Find its volume and the \(z\)-coordinate of its centroid.

Solution

Floor \(z = 0\), ceiling \(z = 1 - x - y\), shadow the triangle \(0\le x\le1\), \(0\le y\le1-x\).

\[ V = \int_0^1\!\!\int_0^{1-x}\!\!\int_0^{1-x-y}dz\,dy\,dx = \int_0^1\!\!\int_0^{1-x}(1 - x - y)\,dy\,dx = \int_0^1\frac{(1-x)^2}{2}\,dx = \frac16 \] \[ \iiint_E z\,dV = \int_0^1\!\!\int_0^{1-x}\frac{(1-x-y)^2}{2}\,dy\,dx = \int_0^1\frac{(1-x)^3}{6}\,dx = \frac{1}{24} \]

\(\bar z = \dfrac{1/24}{1/6} = \dfrac14\). By symmetry \(\bar x = \bar y = \tfrac14\) too — the centroid of a tetrahedron is a quarter of the way from the base to the apex, the 3D analogue of a triangle's one-third rule.

8Cylindrical: energy in a coaxial cable · EE12 pts

Between the conductors of a coaxial cable (inner radius \(a\), outer radius \(b\)) the field is \(\mathbf{E} = \dfrac{V_0}{r\ln(b/a)}\,\hat{\mathbf{r}}\). (a) Find the stored energy per unit length, \(W' = \iiint\tfrac12\varepsilon|\mathbf{E}|^2\,dV\). (b) Using \(W' = \tfrac12C'V_0^2\), deduce the capacitance per unit length and evaluate it for \(a = 1\) mm, \(b = 3.5\) mm, \(\varepsilon = \varepsilon_0 = 8.854\times10^{-12}\) F/m.

Solution

(a) Cylindrical coordinates, unit length in \(z\):

\[ W' = \frac{\varepsilon}{2}\left(\frac{V_0}{\ln(b/a)}\right)^2\int_0^{1}\!\!\int_0^{2\pi}\!\!\int_a^b \frac{1}{r^2}\;r\,dr\,d\theta\,dz = \frac{\varepsilon V_0^2}{2\ln^2(b/a)}\cdot2\pi\cdot\ln\frac{b}{a} = \frac{\pi\varepsilon V_0^2}{\ln(b/a)} \]

The Jacobian \(r\) cancelled one power of the \(1/r^2\), leaving \(\int dr/r = \ln(b/a)\) — the same logarithm that appears in the field.

(b) \(\tfrac12C'V_0^2 = \dfrac{\pi\varepsilon V_0^2}{\ln(b/a)}\) gives

\[ C' = \frac{2\pi\varepsilon}{\ln(b/a)} = \frac{2\pi(8.854\times10^{-12})}{\ln 3.5} = \frac{5.563\times10^{-11}}{1.253}\approx 44\ \text{pF/m} \]

This is the textbook coaxial capacitance formula, derived here purely from a triple integral of the energy density. Real RG-58 cable (polyethylene, \(\varepsilon_r\approx2.3\)) comes in near 100 pF/m — this number times 2.3.

9Spherical coordinates10 pts

Find the centroid of the solid upper half-ball \(x^2+y^2+z^2\le4\), \(z\ge0\).

Solution

By symmetry \(\bar x = \bar y = 0\). The volume is \(\tfrac12\cdot\tfrac43\pi(8) = \tfrac{16\pi}{3}\). For \(\bar z\), use \(z = \rho\cos\phi\):

\[ \iiint_E z\,dV = \int_0^{2\pi}\!\!\int_0^{\pi/2}\!\!\int_0^2 (\rho\cos\phi)\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi\cdot\left[\frac{\rho^4}{4}\right]_0^2\cdot\int_0^{\pi/2}\sin\phi\cos\phi\,d\phi = 2\pi\cdot4\cdot\frac12 = 4\pi \] \[ \bar z = \frac{4\pi}{16\pi/3} = \frac34 \]

In general \(\bar z = \tfrac38R\) for a hemisphere; with \(R = 2\) that is \(0.75\). ✓ Note the limits: \(\phi\) runs only to \(\pi/2\) for the upper half.

10Change of variables10 pts

Evaluate \(\iint_R (x + y)\,e^{x-y}\,dA\) where \(R\) is the region bounded by \(x + y = 1\), \(x + y = 3\), \(x - y = 0\) and \(x - y = 2\).

Solution

Let \(u = x + y\), \(v = x - y\). Then \(R\) becomes the rectangle \(1\le u\le3\), \(0\le v\le2\). Invert: \(x = \tfrac{u+v}{2}\), \(y = \tfrac{u-v}{2}\).

\[ \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix}\tfrac12 & \tfrac12\\ \tfrac12 & -\tfrac12\end{vmatrix} = -\frac14 - \frac14 = -\frac12,\qquad |J| = \frac12 \] \[ \iint_R (x+y)e^{x-y}\,dA = \int_0^2\!\!\int_1^3 u\,e^{v}\cdot\frac12\,du\,dv = \frac12\left[\frac{u^2}{2}\right]_1^3\Big[e^{v}\Big]_0^2 = \frac12\cdot4\cdot(e^2 - 1) = 2(e^2 - 1)\approx 12.8 \]

The absolute value of the Jacobian is what enters — a negative determinant just means the map reverses orientation. The region in \(xy\) is a tilted rectangle of area \(2\times2\times\tfrac12 = 2\); the \(uv\) rectangle has area 4 and the factor \(\tfrac12\) fixes that. ✓

Scoring

Score	Where you are
90–100	Solid. On to vector fields.
75–89	Limits are right, arithmetic is leaking. Evaluate inner integrals on a separate line and substitute the limits explicitly.
60–74	Limit errors — a variable in an outer limit, or a missing Jacobian. Redo each miss starting from a sketch of the region.
< 60	Spend an hour in the region lab until strips and limits are automatic, then retry.

Where the points usually go

The missing \(r\) (Q4, Q8) and \(\rho^2\sin\phi\) (Q9); trying to swap limits without redrawing the region (Q3); integrating \(\phi\) to \(\pi\) instead of \(\pi/2\) for a half-ball (Q9); and forgetting the absolute value on the Jacobian (Q10). Every one of these is caught by checking units or sign of the final answer.