Week 8: Exercises

Exercises – Long Day

Note

The expression \(|\det{\pmb{J}_{\pmb{r}}(\pmb{u})}|\) is in mathematical literature often referred to as “the Jacobian”. This phrase is also often used for the Jacobian matrix, though. To avoid confusion, we will at DTU refer to \(|\det{\pmb{J}_{\pmb{r}}(\pmb{u})}|\) as “the Jacobian function”, although this is not a phrase you will see in literature outside of DTU. Alternatively, we might simply refer to it in full: “the absolute value of the determinant of the Jacobian matrix”.

1: Plane Integrals over Rectangles

Question a

Consider the region \(B=\left\lbrace (x,y) \bigm| 0\leq x\leq 2 \wedge -1\leq y\leq 0\right\rbrace\) in \(\mathbb{R}^2\). Calculate the plane integral

\[\begin{equation*} \int_B (x^2y^2+x) \mathrm{d}\pmb{x} \end{equation*}\]

using the formula for double integrals over (axis-parallel) rectangles.

Question b

Let us calculate the plane integral from above one more time, but now in a manner that at first glance may appear more complicated. Use the change-of-variables theorem for integrals over \(\mathbb{R}^2\).

Hint

First, find a parametrization of the region. Notice that \(B\) is an axis-parallel rectangle in the \((x,y)\) plane.

Hint

A possible parametrization is simply \(\pmb{r}(u,v)=(2u,v-1)\), where \((u,v) \in [0,1]^2\). An alternative is \(\pmb{r}(u,v)=(2u,-v)\), where \((u,v) \in [0,1]^2\). Have a try with both.

Hint

Find the Jacobian function (the absolute value of the Jacobian determinant). Then express the integrand \(x^2y^2+x\) in terms of \(u\) and \(v\).

Hint

Remember the link \((x,y) = \pmb{r}(u,v)\).

Answer

\[\begin{equation*} \int_B x^2y^2+x \;\mathrm{d}\pmb{x}=\frac{26}{9} \end{equation*}\]

Question c

Calculate the plane integral

\[\begin{equation*} \int_B \frac{y}{1+xy} \;\mathrm{d}\pmb{x}, \quad\text{where}\quad B=\left\lbrace (x,y) \mid 0\leq x\leq 1 \, \wedge \, 0\leq y\leq 1\right\rbrace. \end{equation*}\]

Hint

Notice that \(B\) once again is an axis-parallel rectangle in the \((x,y)\) plane.

Hint

\[\begin{equation*} \int_B \left(\frac{y}{1+xy}\right) \;\mathrm{d}\pmb{x}=\int_{0}^{1} \left(\int_{0}^{1} \frac{v}{1+uv} \mathrm{d}u \right)\mathrm{d}v \end{equation*}\]

Hint

With respect to the innermost integral, use integration by substitution where your inner function could be

\[\begin{equation*} t=g(u)=1+uv, \end{equation*}\]

and your outer function could be

\[\begin{equation*} f(t)=\displaystyle{\frac 1t}. \end{equation*}\]

Answer

\[\begin{equation*} \int_B \frac{y}{1+xy} \;\mathrm{d}\pmb{x}=2\ln 2-1 \end{equation*}\]

2: Polar Coordinates

A function \(f:\mathbb{R}^2 \to \mathbb{R}\) is given by

\[\begin{equation*} f(x,y)=x^2-y^2. \end{equation*}\]

For a given point \(\pmb{x}=(x,y)\) in the plane, let \(r = \Vert \pmb{x} \Vert\) denote the distance from the point to the origin \((0,0)\). Also, let \(\theta\) denote the angle between the \(x\)-axis and the position vector to the point - for the sign of the angle \(\theta\), we define the positive angular orientation as counterclockwise (as usual). A set of points \(B\) contains all points that fulfill (in polar coordinates),

\[\begin{equation*} 0\leq r \leq a \, \text{ and } \, -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}, \end{equation*}\]

where \(a\) is an arbitrary positive real number.

Question a

Make a sketch of \(B\), and determine the area of \(B\), first using integration and then purely from elementary geometric considerations.

Hint

You may use this parametric representation of \(B\): \((x,y)=\pmb{p}(u,v)=(u\cos(v),u\sin (v))\) for \(-\frac{\pi}{4} \leq v \leq \frac{\pi}{2}\) and \(0\leq u \leq a\). If you wish, you may of course rename \(u\) to \(r\) and \(v\) to \(\theta\) (which are often-used symbols when dealing with polar coordinates).

Answer

The area is \(\frac{3}{8} a^2\pi\).

Question b

Determine the plane integral \(\int_B f(x,y) \;\mathrm{d}\pmb{x}\).

Hint

For finding antiderivatives the following trigonometric formula for doubled angles may be of use:

\[\begin{equation*} cos^2(v)-\sin^2(v)=\cos(2v). \end{equation*}\]

Answer

\[\begin{equation*} \int_B (x^2-y^2) \;\mathrm{d}\pmb{x}=\frac{a^4}{8} \end{equation*}\]

3: Integration over Circular Ring Segment

A function \(f:\mathbb{R}^2 \setminus \{(0,0)\} \to \mathbb{R}\) is given by

\[\begin{equation*} f(x,y) = \frac{x}{x^2+y^2}. \end{equation*}\]

A closed region \(B\) in the plane is shown on the following figure. The region is a segment of a circular ring centered at the origin.

Question a

Describe \(B\) by reading from the figure the radius \(r\) and the angle \(\theta\) (in radians). This corresponds to determining the integration limits in the following integrals. Then create an integral that calculates the area of \(B\). Finally carry out the calculating of the area of \(B\).

Answer

Limits: \(2 \leq r \leq 5\) and \(\frac{\pi}{2} \leq \theta \leq \frac{7\pi}{4}\). The area is \(\int_B 1 \;\mathrm{d}\pmb{x} = \frac{105\pi}{8}\).

Question b

Argue that \(f\) is Riemann integrable over \(B\). Determine the plane integral \(\int_B f(x,y) \;\mathrm{d}\pmb{x}\).

Hint

Use the fact that \(x^2+y^2 = r^2\) and \(x = r\cos(\theta)\) in polar coordinates. Remember to include the Jacobian determinant \(r\).

Answer

\[\begin{equation*} \int_B \frac{x}{x^2+y^2} \;\mathrm{d}\pmb{x} = -\frac{3\sqrt{2}}{2} - 3 \approx -5.12 \end{equation*}\]

4: Plane Integral with Parametrization I

In the \((x,y)\) plane we are given the point \(P_0=(1,2)\) and the set of points

\[\begin{equation*} C=\left\lbrace (x,y)\Big\vert \frac 32\leq y \leq \frac 52 \wedge 0\leq x\leq \frac 12 y^2\right\rbrace. \end{equation*}\]

Question a

Make a preliminary sketch of \(C\) and provide a parameterization \(\pmb{r}(u,v)\) of \(C\) with appropriate intervals for \(u\) and \(v\), i.e., specify \(\Gamma\) such that \(\pmb{r}(\Gamma) = C\).

Hint

When a region is defined by one variable (here \(y\)) being located in a fixed interval while the other (\(x\)) depends on the first, the easiest is often to let the “free” variable be the first parameter:

1. Let \(u = y\) such that \(u\) runs through the interval \([\frac{3}{2}, \frac{5}{2}]\).

2. For a fixed \(u\), \(x\) must be located between \(0\) and \(\frac{1}{2}u^2\). We can achieve this by letting \(x = v \cdot (\frac{1}{2}u^2)\), where \(v \in [0, 1]\).

3. Now try writing out \(\pmb{r}(u,v) = \begin{bmatrix} x(u,v) \\ y(u,v) \end{bmatrix}\).

Note that you can choose different orders or scalings as long as the region is covered precisely once (with no overlap).

Justify that your chosen parameterization is injective - if it is not injective, then you must find a new one.

Question b

Determine the two parameter values \(u_0\) and \(v_0\) such that \(\pmb{r}(u_0,v_0)=P_0\). Make an illustration of \(C\) (a hand-drawn sketch or a plot in Sympy) where you from \(P_0\) draw the tangent vectors \(\pmb{r}'_u(u_0,v_0)\) and \(\pmb{r}'_v(u_0,v_0)\). Determine the area of the parallelogram that is spanned by these tangentvectors (more on this in the Short Day Exercise 2: The Volume of a Parallelotope).

Question c

Determine the Jacobian determinen that corresponds to \(\pmb{r}(u,v)\), and argue that the two column vectors that constitute the Jacobian matrix are linearly independent for all \((u,v) \in \Gamma\). Calculate the Jacobian determinant at the point \((u_0,v_0)\).

Question d

Calculate the plane integral

\[\begin{equation*} \int_C \frac{1}{y^2+x} \mathrm{d}\pmb{x} \end{equation*}\]

using the change-of-variables theorem for integrals over \(\mathbb{R}^2\). You must argue that changing variables is a usable method for this case. Check your result with the theorem on axis-parallel regions.

Answer

The parametrization can for example be \(\pmb{r}(u,v)=(\tfrac{1}{2}vu^2,u)\), where \(u\in\left[ \tfrac 32,\tfrac 52\right]\) and \(v\in\left[ 0,1\right]\). The corresponding Jacobian determinant is then \(-\frac{1}{2}u^2\).

\[\begin{equation*} \int_C \frac{1}{x^2+y} \mathrm{d}\pmb{x}=\ln(3)-\ln(2). \end{equation*}\]

5: The Plane Integral with Parametrization II

We want to determine the plane integral

\[\begin{equation*} \int_B 2xy\,\mathrm{d} \pmb{x}, \end{equation*}\]

where \(B=\pmb{r}([0,1]^2)\) is given by the parametrization

\[\begin{equation*} \pmb{r}(u,v)=(u,v(1-u)),\;\text{where}\; u\in\left[ 0,1\right]\text{ and } v\in\left[ 0,1\right]. \end{equation*}\]

Follow the below steps.

Question a

Describe the region \(B\) using inequalities (such as \(x+5y\ge 7\)). Then sketch \(B\).

Hint

Fix \(u\) and you will see that \(\pmb{r}(u,v)\) runs through a line as \(v\) is varied.

Answer

The region \(B\) can easily be sketched based on the following inequalities:

\[\begin{equation*} B=\left\lbrace (x,y) \mid 0\leq x \, \wedge \, 0\leq y, x+y\leq 1\right\rbrace. \end{equation*}\]

Question b

Determine the Jacobian determinant for the parametrization \(\pmb{r}(u,v)\). Is the Jacobian determinant different from zero on the interior of the parameter domain (this is a requirement for using the change-of-variables theorem)?

Answer

\(\mathrm{det}(\pmb{J}_{\pmb{r}}(u,v)) = 1-u > 0\). The Jacobian determinant is positive for \(u,v \in ]0,1[\), so on the interior of the parameter domain \(\Gamma^\circ\), where \(\Gamma = [0,1]^2\).

Question c

Now determine the wanted integral.

Hint

You must substitute the first and second coordinates of the parametrization into the function \(2xy,\) in place of \(x\) and \(y\), respectively, and then multiply by (the absolute value of) the Jacobian determinant. Then you must do the integration, first with respect to \(u\), then to \(v\).

Hint

You must integrate

\[\begin{equation*} \int_0^1\left(\int_0^1 2u^3v-4u^2v+2uv\;\mathrm{d}u\right)\mathrm{d}v. \end{equation*}\]

Hint

An antiderivative with respect to \(u\) is:

\[\begin{equation*} {\frac 12u^4v-\frac 43u^3v+u^2v}. \end{equation*}\]

When you substitute in the \(u\) limits, you get \(\displaystyle{\frac 16 v}\) which you now must integrate with respect to \(v\).

Answer

\[\begin{equation*} \int_B 2xy \mathrm{d} \pmb{x}=\frac{1}{12} \end{equation*}\]

6: A Triple Integral

Calculate the triple integral

\[\begin{equation*} \displaystyle{\int_1^2\int_1^2\int_1^2 \frac{xy}{z} \mathrm dx\mathrm dy\mathrm dz.}\ \end{equation*}\]

Hint

We integrate over an axis-parallel box!

Answer

\(\displaystyle{\frac 94 \ln(2).}\)

7: An Indefinite Integral in the Plane

Let \(B\) be the unit square \([0,1]^2\). We will in this exercise investigate the indefinite plane integral:

\[\begin{equation*} I = \int_B \frac{1}{x_2-x_1-1} \mathrm{d}\pmb{x}. \end{equation*}\]

The integrand \(f(x_1,x_2)=\frac{1}{x_2-x_1-1}\) is not Riemann integrable over \(B\), since \(f\) is not defined at the point \((x_1,x_2)=(0,1)\). We wish to find out whether it is still possible to give the integral a value by considering limits.

Question a

Find those points in the \((x,y)\) plane where \(f(x_1,x_2)\) is not defined. Find the range of \(f\) as a function on \(B \setminus \{(0,1)\}\).

Hint

Is \(f\) positive or negative on \(B\)?

Question b

Let \(B_a = [a,1] \times [0,1]\) for a fixed \(a \in [0,1]\). Make a sketch of \(B_a\) and create a parametrization of \(B_a\). Determine the Jacobian determinant of the parametrization.

Question c

Calculate the Riemann integral

\[\begin{equation*} I_a := \int_{B_a} \frac{1}{x_2-x_1-1} \mathrm{d}\pmb{x} \end{equation*}\]

for every \(a \in ]0,1]\).

Question d

Calculate the limit of \(I_a\) for \(a \to 0\).

Answer

\(I = -2 \ln(2)\)

Question e

Let \(B_b = [0,1] \times [0,b]\). Define \(I_b := \int_{B_b} \frac{1}{x_2-x_1-1} \mathrm{d}\pmb{x} \). Find \(\lim_{b \to 1} I_b\). Compare with the above.

Exercises – Short Day

1: Parametrized Spatial Region

A region \(B\) in \((x,y,z)\) space is given by the parametric representation

\[\begin{equation*} \pmb{r}(u,v,w)=\big(\frac{1}{2}u^2-v^2,-uv,w\big),\quad u\in \left[ 0,2\right],v\in \left[ 0,2\right],w\in \left[ 0,2\right]. \end{equation*}\]

Question a

In \(B\) we are given the point

\[\begin{equation*} \pmb{x}_0=\pmb{r}(1,1,1). \end{equation*}\]

Find \(\pmb{x}_0\). When placed at \(\pmb{x}_0\), the tangent vectors \(\pmb{r}_u'(1,1,1),\pmb{r}_v'(1,1,1)\) and \(\pmb{r}_w'(1,1,1)\) span a parallelepiped \(P\), see Exercise 2: The Volume of a Parallelotope. Determine the volume of this parallelepiped. It would be good training to illustrate this with Sympy.

Hint

The volume is found as the absolute value of the determinant af the matrix whose columns are the spanning vectors.

Answer

\(\pmb{x}_0=(-1/2, -1, 1)\). The volume of \(P\) is \(3\).

Question b

Determine the absolute value of the Jacobian determinant corresponding to \(\pmb{r}\). Evaluate it at \(\pmb{x}_0\).

Hint

Remember that the Jacobian determinant is the determinanten of the Jacobian matrix. How is this matrix related to what we found in the previous question?

Answer

\(|\pmb{J}_{\pmb{r}}(u,v,w)|=u^2+2v^2\). Altså er \(|\pmb{J}_{\pmb{r}}(1,1,1)|=1^2+2\cdot 1^2 = 3\).

Question c

Calculate the volume of \(B\).

Hint

The volume of \(B\) can be calculated as the integral over the function 1 - and don’t forget the absolute value of the Jacobian determinant as the integrand.

Answer

\(32\).

2: The Volume of a Parallelotope

A parallelotope \(P\) in \(\mathbb{R}^n\) “spanned by” the vectors \(\pmb{a}_1, \pmb{a}_2, \dots, \pmb{a}_n\) is defined by:

\[\begin{equation*} P = \left\{ \pmb{y} \in \mathbb{R}^n \mid \, \pmb{y} = A\pmb{x}, \quad \text{where } x_i \in [0,1] \text{ for }$i=1,2,\dots, n$ \right\}, \end{equation*}\]

where \(A = [\pmb{a}_1 | \pmb{a}_2 | \cdots | \pmb{a}_n]\) is the \(n \times n\) matrix whose \(i\)’th column is \(\pmb{a}_i\). This set of points can in short-hand notation be written as \(P=A([0,1]^n)\).

It can be shown with tools solely from Mathematics 1a (in particular the characterization of the determinant) that the \(n\)-dimensional volume of \(P\) is:

\[\begin{equation*} \mathrm{vol}_n(P) = |\mathrm{det}(A)|. \end{equation*}\]

(For the interested student, such a proof can be found here https://textbooks.math.gatech.edu/ila/determinants-volumes.html)

In \(\mathbb{R}^2\), a parallelotope is the well-known pallelogram, and \(\mathrm{vol}_n(P)\) is then the area of \(P\), while it in \(\mathbb{R}^3\) becomes a parallelepiped with a volume.

Question a

Show that \(\mathrm{vol}_n(P) = |\mathrm{det}(A)|\) using the change-of-variables theorem for integrals over \(\mathbb{R}^n\).

Hint

Since \(P=A([0,1]^n)\), we will choose \(\Gamma = [0,1]^n\). What is the expression of the corresponding parametrization?

Hint

\(\pmb{r}(\pmb{u}) = A \pmb{u}\), where \(\pmb{u} \in \Gamma = [0,1]^n\). What is the corresponding Jacobian matrix?

Answer

The Jacobian determinant is \(\mathrm{det}(A)\). For computation of \(\mathrm{vol}_n(P)\), we need to use \(f(\pmb{x}) = 1\). The full integrand is thus the constant \(|\mathrm{det}(A)|\), which is to be integrated over \(\Gamma = [0,1]^n\). This just becomes \(|\mathrm{det}(A)|\). The integral we have calculated is by definition equal to \(\mathrm{vol}_n(P)\).

In the rest of this exercise we want to investigate the proposition \(\mathrm{vol}_n(P) = |\mathrm{det}(A)|\) without use of integration techniques.

Question b

Let \(n=2\). Choose two linearly independent vectors \(\pmb{a}_1, \pmb{a}_2\) in \(\mathbb{R}^2\). It might be smart to choose \(\pmb{a}_1 \in \mathrm{span}(\pmb{e}_1)\). Calculate (using elementary geometric considerations) the area of the parallelogram “spanned by” the two vectors. Also calculate \(|\mathrm{det}(A)|\) and compare the results.

Hint

You may need to do an orthogonal projection of \(\pmb{a}_2\) on the orthogonal complement to \(\pmb{a}_1\).

Question c

Let \(n=2\), and now let \(\pmb{a}_1, \pmb{a}_2\) be arbitrary but linearly independent vectors in \(\mathbb{R}^2\). Can you prove the formula \(\mathrm{area}(P) = |\mathrm{det}(A)|\), where \(P\) is the parallelogram “spanned by” the two vectors? You may assume (why?) that \(\pmb{a}_1 \in \mathrm{span}(\pmb{e}_1)\), if this helps in your argumentation.

Hint

You might again need to do the orthogonal projection of \(\pmb{a}_2\) on the orthogonal complement to \(\pmb{a}_1\).

Answer

Below is given a proof without use of trigonometric identities:

As rotation does not change the area of a region, then by rotating the parallelogram we can assume that \(\pmb{a}_1 \in \mathrm{span}(\pmb{e}_1)\), meaning that \(A\) is an upper triangular matrix \(A = [a_{i,j}]\) with \(a_{1,2}=0\). As \(A\) is an upper triangular matrix, its determinant is the product of its diagonal elements, so \(\mathrm{det}(A) = a_{1,1} a_{2,2}\). Hence, \(|\mathrm{det}(A)| = |a_{1,1} a_{2,2}|\). We now just need to prove that this i also equal to the area.

The area of the parallelogram is given as the length of \(\pmb{a}_1\) (meaning \(|a_{1,1}|\)) multiplied by “the height”, which is the length of the orthogonal projection of \(\pmb{a}_2\) on \(\mathrm{span}(\pmb{e}_2)\) (so, \(|\langle \pmb{a}_2, \pmb{e}_2 \rangle| = |a_{2,2}|\)). Hence, the area is: \(|a_{1,1}| |a_{2,2}| = |a_{1,1} a_{2,2}|\).

Question d

Let \(n=3\). Choose three linearly independent vectors \(\pmb{a}_1, \pmb{a}_2, \pmb{a}_3\) in \(\mathbb{R}^3\). It can be smart to choose \(\pmb{a}_1, \pmb{a}_2 \in \mathrm{span}(\pmb{e}_1, \pmb{e}_2)\). Calculate (using elementary geometric considerations) the volume of the parallelepiped “spanned by” the three vectors. Also calculate \(|\mathrm{det}(A)|\) and compare the two results.

Hint

You may need to do an orthogonal projection of \(\pmb{a}_2\) on the orthogonal complement to \(\pmb{a}_1\).

Question e (Extra, can Wait Until After the Exercises of the Day)

Let \(n=3\), and now let \(\pmb{a}_1, \pmb{a}_2, \pmb{a}_3\) be arbitrary but linearly independent vectors in \(\mathbb{R}^3\). Can you prove the formula \(\mathrm{areal}(P) = |\mathrm{det}(A)|\), where \(P\) is the parallelepiped “spanned by” the three vectors? You may assume (why?) that \(\pmb{a}_1, \pmb{a}_2 \in \mathrm{span}(\pmb{e}_1, \pmb{e}_2)\), if that helps your argumentation.

3: Mass Distributions in the \((x,y)\) Plane

Consider the sets of points in \(\mathbb{R}^2\) given by:

\[\begin{equation*} B=\left\lbrace (x,y) \in \mathbb{R}^2 \;\Big\vert \; 1\leq x\leq 2 \, \wedge \, 0\leq y\leq x^3\right\rbrace, \end{equation*}\]

and consider (again)

\[\begin{equation*} C=\left\lbrace (x,y) \in \mathbb{R}^2 \;\Big\vert \; \frac 32\leq y \leq \frac 52 \wedge 0\leq x\leq \frac 12 y^2\right\rbrace. \end{equation*}\]

We will think of \(f(x,y)\) as a function that expresses the mass density at the point \((x,y)\) (so, with units such as \(\mathrm{kg/m^2}\)).

Question a

Assume that the mass density is constant, \(f(x,y)=1\) for \((x,y)\in B\). Determine the mass and centre of mass of \(B\).

Hint

The mass is \(M = \int_B 1 \mathrm{d}(x,y)\). Find the centre-of-mass formula in the textbook in the section about integration of vector functions.

Answer

The mass is \(\frac{15}{4}\) and the centre of mass is \((x,y)=(\frac{124}{75},\frac{254}{105})\).

Question b

Assume that the mass density is \(f(x,y)=x^2\) for \((x,y)\in B\). Determine the mass and the centre of mass of \(B\).

Answer

The mass is \(\frac{21}{2}\) and the centre of mass is \((x,y)=(\frac{254}{147},\frac{73}{27})\).

Question c

Assume that the mass density is constant \(f(x,y)=1\) for \((x,y)\in C\). Determine the mass and the centre of mass of \(C\).

Answer

The mass is \(\frac{49}{24}\) and the centre of mass is \((x,y)=(\frac{4323}{3920},\frac{102}{49})\).

Question d

Assume that the mass density is \(f(x,y)=x^2\) for \((x,y)\in C\). Determine the mass and the centre of mass of \(C\).

Answer

The mass is \(\frac{37969}{10752}\) and the centre of mass is \((x,y)=(\frac{6767047}{3645024},\frac{84014}{37969})\).

4: Spherical Regions in 3D Space

Consider the spatial region \(\pmb{r}(\Gamma)\) given by

\[\begin{equation*} \pmb{r}(u,v,w)=\big(u\sin(v)\cos(w),u\sin(v)\sin(w),u\cos(v)\big), \quad (u,v,w) \in \Gamma, \end{equation*}\]

where \(\Gamma = [a,b] \times [c,d] \times [e,f] \subset [0, \infty[ \times [0,\pi] \times [0,2\pi]\). Meaning, we are dealing with the following parameter values: \(u\in [a,b],v\in [c,d],w\in [e,f]\).

Question a

What do the parameters represent?

Question b

Let \(A\) be the region that is determined by the choice:

\[\begin{equation*} a=1,b=3,c=\frac{\pi}{4},d=\frac{\pi}{3},e=0,f=\frac{3\pi}{4}, \end{equation*}\]

and let \(B\) be the region determined by the choice:

\[\begin{equation*} a=2,b=4,c=\frac{\pi}{4},d=\frac{\pi}{2},e=-\frac{\pi}{4},f=\frac{\pi}{4}. \end{equation*}\]

Describe in words each of the regions \(A\), \(B\) and \(A\cap B\), and calculate their volumes.

Answer

\(\mathrm{vol} (A)=\frac{13\pi(\sqrt 2-1)}{4}\).

\(\mathrm{vol} (B)=\frac{14\pi\sqrt 2}{3}\).

\(\mathrm{vol} (A\cap B)=\frac{19\pi(\sqrt 2-1)}{24}\).

By the way, \(\mathrm{vol}(A\cup B)\) can be calculated as \(\mathrm{vol}(A) + \mathrm{vol}(B) - \mathrm{vol}(A\cap B)\).

Question c

Let \(\boldsymbol x=(x_1,x_2,x_3)\). Calculate all of the integrals

\[\begin{equation*} \int_A x_1 \, \mathrm{d}\pmb{x}, \quad \int_Bx_1 \, \mathrm{d}\pmb{x} \quad \text{and} \quad \int_{A\cap B}x_1 \, \mathrm{d}\pmb{x}. \end{equation*}\]

Answer

\[\begin{equation*} \int_A x_1 \, \mathrm{d}\pmb{x}=\frac{5\sqrt 2}{2}+\frac{5\pi\sqrt 2}{12}-\frac{5\sqrt 2\sqrt 3}{4} \end{equation*}\]

\[\begin{equation*} \int_B x_1 \, \mathrm{d}\pmb{x}=\frac{15\pi\sqrt 2}{2}+15\sqrt 2 \end{equation*}\]

\[\begin{equation*} \int_{A\cap B}x_1 \, \mathrm{d}\pmb{x}=\frac{65\sqrt 2}{32}+\frac{65\pi\sqrt 2}{192}-\frac{65\sqrt 2\sqrt 3}{64} \end{equation*}\]