Print preview

Highlight workspace

Worksheet Assigned Problems for Section 15.4

The problems listed below are assigned to be included in your problem set portfolio. Note that a specific selection of these problems will also form the written homework assignments. I recommend working through all of them to ensure a solid grasp of the material. Reach out to Richard for help if you get stuck or have any questions.

🔗

The solutions will be posted after the written homework due dates. If you have any questions about your work, talk to Richard and he is happy to discuss the process with you.

🔗

15.4.3.

Sketch the region \(\c{D} = \left\{ (x,y) \in \R^2 \mid x \geq 0, \, y \geq 0 \, \text{and } x^2 + y^2 \leq 4 \right\}\) and integrate \(f(x,y) = xy\) over \(\c{D}\) using polar coordinates.

🔗

Solution.

The region \(\c{D}\) is bounded by \(x \geq 0\) (the right half-plane), \(y \geq 0\) (the upper half-plane), and \(x^2 + y^2 \leq 4\) (a disk of radius 2). This describes exactly the quarter-circle of radius 2 in the first quadrant.

🔗

In polar coordinates, this region is defined by the limits:

\begin{align*} 0 \amp \leq r \leq 2 \\ 0 \amp \leq \theta \leq \frac{\pi}{2} \end{align*}

🔗

We convert the integrand \(f(x,y) = xy\) using \(x = r\cos\theta\) and \(y = r\sin\theta\text{.}\) This gives \(f(r,\theta) = (r\cos\theta)(r\sin\theta) = r^2\cos\theta\sin\theta\text{.}\) Remembering to include the extra \(r\) from the differential \(dA = r\,dr\,d\theta\text{,}\) we set up the iterated integral:

\begin{align*} \iint_\c{D} xy \, dA \amp= \int_0^{\pi/2} \int_0^2 (r^2\cos\theta\sin\theta) r \, dr \, d\theta \\ \amp= \int_0^{\pi/2} \int_0^2 r^3\cos\theta\sin\theta \, dr \, d\theta \end{align*}

🔗

Evaluate the inner integral with respect to \(r\text{:}\)

\begin{align*} \int_0^2 r^3\cos\theta\sin\theta \, dr \amp= \left[ \frac{1}{4}r^4\cos\theta\sin\theta \right]_{r=0}^{r=2} \\ \amp= \frac{1}{4}(16)\cos\theta\sin\theta - 0 = 4\cos\theta\sin\theta \end{align*}

🔗

Substitute this into the outer integral and evaluate with respect to \(\theta\text{.}\) We can use the double angle identity \(2\sin\theta\cos\theta = \sin(2\theta)\) to rewrite \(4\cos\theta\sin\theta\) as \(2\sin(2\theta)\text{:}\)

\begin{align*} \int_0^{\pi/2} 2\sin(2\theta) \, d\theta \amp= \left[ -\cos(2\theta) \right]_0^{\pi/2} \\ \amp= -\cos\left(2\cdot\frac{\pi}{2}\right) - (-\cos(0)) \\ \amp= -\cos(\pi) + \cos(0) = -(-1) + 1 = 2 \end{align*}

🔗

15.4.13.

For the following integral, sketch the region of integration and evaluate by changing to polar coordinates.

\begin{equation*} \int_{-1}^2 \int_0^\sqrt{4 - x^2} \lp x^2 + y^2 \rp \, dy \, dx \end{equation*}

🔗

Solution.

Let’s analyze the given limits of integration:

\begin{align*} -1 \amp \leq x \leq 2 \\ 0 \amp \leq y \leq \sqrt{4 - x^2} \end{align*}

The \(y\)-bounds tell us the region lies in the upper half-plane (\(y \geq 0\)) and is bounded above by the semicircle \(y = \sqrt{4 - x^2}\text{,}\) which corresponds to a circle of radius 2. However, \(x\) only goes from \(-1\) to \(2\text{,}\) meaning we do not have the full upper half-circle (which would go from \(x = -2\) to \(x = 2\)).

🔗

In polar coordinates, the radius clearly goes from \(r = 0\) to \(r = 2\text{.}\) The angle \(\theta\) starts at the positive \(x\)-axis where \(x = 2\) (so \(\theta = 0\)). It sweeps counterclockwise until it hits the point on the circle where \(x = -1\text{.}\) Using \(x = r\cos\theta\text{,}\) we have \(-1 = 2\cos\theta \implies \cos\theta = -\frac{1}{2}\text{.}\) In the upper half-plane, this corresponds to the angle \(\theta = \frac{2\pi}{3}\text{.}\) Thus, our bounds are \(0 \leq r \leq 2\) and \(0 \leq \theta \leq \frac{2\pi}{3}\text{.}\)

🔗

Converting the integrand, we know \(x^2 + y^2 = r^2\text{.}\) Including the polar differential \(dA = r\,dr\,d\theta\text{,}\) the integral becomes:

\begin{gather*} \int_0^{2\pi/3} \int_0^2 (r^2) r \, dr \, d\theta = \int_0^{2\pi/3} \int_0^2 r^3 \, dr \, d\theta \end{gather*}

🔗

Evaluate the inner integral with respect to \(r\text{:}\)

\begin{align*} \int_0^2 r^3 \, dr \amp= \left[ \frac{1}{4}r^4 \right]_0^2 = \frac{1}{4}(16) - 0 = 4 \end{align*}

🔗

Substitute this into the outer integral and evaluate with respect to \(\theta\text{:}\)

\begin{align*} \int_0^{2\pi/3} 4 \, d\theta \amp= \Big[ 4\theta \Big]_0^{2\pi/3} = 4\left(\frac{2\pi}{3}\right) - 0 = \frac{8\pi}{3} \end{align*}

🔗

15.4.19.

Calculate the integral of \(f(x,y) = x - y\) over the region \(\c{D} = \left\{ (x,y) \in \R^2 \mid x^2 + y^2 \leq 1 \text{ and } x + y \geq1\right\}\) by changing to polar coordinates.

🔗

Solution.

The region \(\c{D}\) is bounded by the circle \(x^2 + y^2 = 1\) and the line \(x + y = 1\text{.}\) These two curves intersect at \((1,0)\) and \((0,1)\text{,}\) so the region lies entirely in the first quadrant, making our angular bounds \(0 \leq \theta \leq \frac{\pi}{2}\text{.}\)

🔗

For a fixed angle \(\theta\text{,}\) a ray from the origin enters the region at the line and exits at the circle.

Outer bound (circle): \(x^2 + y^2 = 1 \implies r = 1\)

🔗
Inner bound (line): \(x + y = 1 \implies r\cos\theta + r\sin\theta = 1 \implies r(\cos\theta + \sin\theta) = 1 \implies r = \frac{1}{\cos\theta + \sin\theta}\)

🔗

🔗

The integrand converts to \(x - y = r\cos\theta - r\sin\theta = r(\cos\theta - \sin\theta)\text{.}\) With \(dA = r\,dr\,d\theta\text{,}\) the integral is:

\begin{gather*} \int_0^{\pi/2} \int_{\frac{1}{\cos\theta + \sin\theta}}^1 \big(r(\cos\theta - \sin\theta)\big) r \, dr \, d\theta = \int_0^{\pi/2} \int_{\frac{1}{\cos\theta + \sin\theta}}^1 r^2(\cos\theta - \sin\theta) \, dr \, d\theta \end{gather*}

🔗

Evaluate the inner integral with respect to \(r\text{:}\)

\begin{align*} \int_{\frac{1}{\cos\theta + \sin\theta}}^1 r^2(\cos\theta - \sin\theta) \, dr \amp= \left[ \frac{1}{3}r^3(\cos\theta - \sin\theta) \right]_{r=\frac{1}{\cos\theta + \sin\theta}}^{r=1} \\ \amp= \frac{1}{3}(1)^3(\cos\theta - \sin\theta) - \frac{1}{3}\left(\frac{1}{\cos\theta + \sin\theta}\right)^3(\cos\theta - \sin\theta) \\ \amp= \frac{1}{3}(\cos\theta - \sin\theta) - \frac{1}{3}\frac{\cos\theta - \sin\theta}{(\cos\theta + \sin\theta)^3} \end{align*}

🔗

Substitute this into the outer integral:

\begin{gather*} \int_0^{\pi/2} \left( \frac{1}{3}(\cos\theta - \sin\theta) - \frac{1}{3}\frac{\cos\theta - \sin\theta}{(\cos\theta + \sin\theta)^3} \right) \, d\theta \end{gather*}

We can split this into two integrals and evaluate. For the first part:

\begin{align*} \int_0^{\pi/2} \frac{1}{3}(\cos\theta - \sin\theta) \, d\theta \amp= \frac{1}{3} \Big[ \sin\theta + \cos\theta \Big]_0^{\pi/2} \\ \amp= \frac{1}{3}\big(\sin(\pi/2) + \cos(\pi/2)\big) - \frac{1}{3}\big(\sin(0) + \cos(0)\big) \\ \amp= \frac{1}{3}(1 + 0) - \frac{1}{3}(0 + 1) = 0 \end{align*}

For the second part, let \(u = \cos\theta + \sin\theta\text{.}\) Then \(du = (-\sin\theta + \cos\theta)\,d\theta = (\cos\theta - \sin\theta)\,d\theta\text{.}\) Let’s look at the bounds: when \(\theta = 0\text{,}\) \(u = 1+0=1\text{.}\) When \(\theta = \pi/2\text{,}\) \(u = 0+1=1\text{.}\) Since the upper and lower limits of integration are the same (\(u\) goes from 1 to 1), the definite integral evaluates exactly to 0!

🔗

Therefore, the total integral is \(0 - 0 = 0\text{.}\) (This makes perfect sense geometrically, as the region is symmetric across the line \(y = x\text{,}\) but the function \(x - y\) has perfectly opposite values across that same line, causing everything to cancel out!).

🔗

15.4.23.

Evaluate \(\ds \iint_\c{D} \sqrt{x^2 + y^2} \, dA\text{,}\) where \(\c{D}\) is the domain in the following figure.

🔗

Hint.

Find the equation of the inner circle in polar coordinates and treat the right and left parts of the region separately.

🔗

Solution.

Looking at the figure, the domain \(\c{D}\) is bounded by the outer circle \(x^2 + y^2 = 4\) (which has a radius of 2), and it contains an empty "hole." The hole is a circle centered at \((1, 0)\) with a radius of 1.

🔗

Let’s find the polar equation for the inner boundary. The Cartesian equation is \((x - 1)^2 + y^2 = 1\text{.}\) Expanding this yields:

\begin{align*} x^2 - 2x + 1 + y^2 \amp= 1 \\ x^2 + y^2 \amp= 2x \end{align*}

Substituting our polar conversions (\(x^2 + y^2 = r^2\) and \(x = r\cos\theta\)):

\begin{align*} r^2 \amp= 2r\cos\theta \implies r = 2\cos\theta \end{align*}

🔗

Notice that the inner circle \(r = 2\cos\theta\) only exists in the right half-plane (where \(-\pi/2 \leq \theta \leq \pi/2\)). In the left half-plane (where \(\pi/2 \leq \theta \leq 3\pi/2\)), there is no hole, so the region goes all the way from the origin \(r = 0\) to the outer edge \(r = 2\text{.}\) Therefore, we must split our region into two parts: a left half and a right half.

🔗

The integrand is \(\sqrt{x^2+y^2} = r\text{.}\) With the polar differential \(dA = r\,dr\,d\theta\text{,}\) the function we are integrating becomes \(r \cdot r = r^2\text{.}\)

\begin{gather*} \iint_\c{D} \sqrt{x^2+y^2} \, dA = \int_{\pi/2}^{3\pi/2} \int_0^2 r^2 \, dr \, d\theta + \int_{-\pi/2}^{\pi/2} \int_{2\cos\theta}^2 r^2 \, dr \, d\theta \end{gather*}

🔗

Part 1: The Left Half

\begin{align*} \int_{\pi/2}^{3\pi/2} \int_0^2 r^2 \, dr \, d\theta \amp= \int_{\pi/2}^{3\pi/2} \left[ \frac{1}{3}r^3 \right]_0^2 \, d\theta \\ \amp= \int_{\pi/2}^{3\pi/2} \frac{8}{3} \, d\theta \\ \amp= \left[ \frac{8}{3}\theta \right]_{\pi/2}^{3\pi/2} = \frac{8}{3}\left(\frac{3\pi}{2}\right) - \frac{8}{3}\left(\frac{\pi}{2}\right) = \frac{8\pi}{3} \end{align*}

🔗

Part 2: The Right Half

\begin{align*} \int_{-\pi/2}^{\pi/2} \int_{2\cos\theta}^2 r^2 \, dr \, d\theta \amp= \int_{-\pi/2}^{\pi/2} \left[ \frac{1}{3}r^3 \right]_{2\cos\theta}^2 \, d\theta \\ \amp= \int_{-\pi/2}^{\pi/2} \left( \frac{8}{3} - \frac{8}{3}\cos^3\theta \right) \, d\theta \end{align*}

We separate this into two integrals:

\begin{gather*} \int_{-\pi/2}^{\pi/2} \frac{8}{3} \, d\theta - \frac{8}{3} \int_{-\pi/2}^{\pi/2} \cos^3\theta \, d\theta \end{gather*}

The first integral is simply \(\frac{8}{3}(\pi) = \frac{8\pi}{3}\text{.}\) For the second integral, we use the Pythagorean identity \(\cos^2\theta = 1 - \sin^2\theta\) to set up a \(u\)-substitution where \(u = \sin\theta\) and \(du = \cos\theta \, d\theta\text{.}\) When \(\theta = -\pi/2\text{,}\) \(u = -1\text{.}\) When \(\theta = \pi/2\text{,}\) \(u = 1\text{.}\)

\begin{align*} \int_{-\pi/2}^{\pi/2} \cos^2\theta \cos\theta \, d\theta \amp= \int_{-1}^1 (1 - u^2) \, du \\ \amp= \left[ u - \frac{1}{3}u^3 \right]_{-1}^1 \\ \amp= \left( 1 - \frac{1}{3} \right) - \left( -1 - \left(-\frac{1}{3}\right) \right) = \frac{2}{3} - \left(-\frac{2}{3}\right) = \frac{4}{3} \end{align*}

So the right half evaluates to \(\frac{8\pi}{3} - \frac{8}{3}\left(\frac{4}{3}\right) = \frac{8\pi}{3} - \frac{32}{9}\text{.}\)

🔗

Total Area Summing the left half and the right half:

\begin{align*} \iint_\c{D} \sqrt{x^2+y^2} \, dA \amp= \frac{8\pi}{3} + \left( \frac{8\pi}{3} - \frac{32}{9} \right) \\ \amp= \frac{16\pi}{3} - \frac{32}{9} \end{align*}

🔗

15.4.29.

Use cylindrical coordinates to evaluate \(\ds \iiint_\c{W} x \, dV\) where \(\c{W} = \left\{ (x,y,z) \in \R^3 \mid x^2 + y^2 \leq 16 , \, x \geq 0, \, y \geq 0, \, -3 \leq z \leq 3 \right\}\)

🔗

Solution.

The region \(\c{W}\) is bounded by the cylinder \(x^2 + y^2 = 16\) (radius 4) in the first quadrant (\(x \geq 0\text{,}\) \(y \geq 0\)), between the planes \(z = -3\) and \(z = 3\text{.}\)

🔗

In cylindrical coordinates, this region translates beautifully to constant bounds:

\begin{align*} 0 \amp \leq r \leq 4 \\ 0 \amp \leq \theta \leq \frac{\pi}{2} \\ -3 \amp \leq z \leq 3 \end{align*}

🔗

The integrand is \(x = r\cos\theta\text{.}\) With \(dV = r\,dz\,dr\,d\theta\text{,}\) the integral is:

\begin{align*} \iiint_\c{W} x \, dV \amp= \int_0^{\pi/2} \int_0^4 \int_{-3}^3 (r\cos\theta) r \, dz \, dr \, d\theta \\ \amp= \int_0^{\pi/2} \int_0^4 \int_{-3}^3 r^2\cos\theta \, dz \, dr \, d\theta \end{align*}

🔗

Evaluate the inner integral with respect to \(z\text{:}\)

\begin{align*} \int_{-3}^3 r^2\cos\theta \, dz \amp= \Big[ r^2z\cos\theta \Big]_{z=-3}^{z=3} \\ \amp= r^2(3)\cos\theta - r^2(-3)\cos\theta = 6r^2\cos\theta \end{align*}

🔗

Evaluate the middle integral with respect to \(r\text{:}\)

\begin{align*} \int_0^4 6r^2\cos\theta \, dr \amp= \left[ 2r^3\cos\theta \right]_0^4 \\ \amp= 2(64)\cos\theta - 0 = 128\cos\theta \end{align*}

🔗

Evaluate the outer integral with respect to \(\theta\text{:}\)

\begin{align*} \int_0^{\pi/2} 128\cos\theta \, d\theta \amp= \Big[ 128\sin\theta \Big]_0^{\pi/2} \\ \amp= 128\sin(\pi/2) - 128\sin(0) = 128(1) - 0 = 128 \end{align*}

🔗

15.4.35.

Express the triple integral in cylindrical coordinates.

\begin{equation*} \int_{x = -1}^1 \int_{y = 0}^\sqrt{1 - x^2} \int_{z = 0}^{x^2 + y^2} f(x,y,z) \, dz \, dy \, dx \end{equation*}

🔗

Solution.

We need to translate the rectangular bounds into cylindrical bounds. The given limits are:

\begin{align*} -1 \amp \leq x \leq 1 \\ 0 \amp \leq y \leq \sqrt{1 - x^2} \\ 0 \amp \leq z \leq x^2 + y^2 \end{align*}

🔗

The \(z\)-bounds are \(0 \leq z \leq x^2 + y^2\text{.}\) In cylindrical coordinates, \(x^2 + y^2 = r^2\text{,}\) so our new \(z\)-bounds are \(0 \leq z \leq r^2\text{.}\)

🔗

The \(x\) and \(y\) bounds describe the projection of the solid on the \(xy\)-plane. \(y\) ranges from \(0\) to the upper semicircle \(y = \sqrt{1 - x^2}\) (which is \(x^2 + y^2 = 1\)). \(x\) sweeps completely from \(-1\) to \(1\text{.}\) This outlines the top half of the unit disk. In cylindrical (polar) coordinates, this is \(0 \leq r \leq 1\) and \(0 \leq \theta \leq \pi\text{.}\)

🔗

Substituting \(x = r\cos\theta\) and \(y = r\sin\theta\) into the function, and adding the \(r\) for the volume differential \(dV = r\,dz\,dr\,d\theta\text{,}\) the integral becomes:

\begin{gather*} \int_0^\pi \int_0^1 \int_0^{r^2} f(r\cos\theta, r\sin\theta, z) r \, dz \, dr \, d\theta \end{gather*}

🔗

15.4.39.

Find the volume of the region appearing between the two surfaces in the following figure.

🔗

Solution.

Looking at the figure, the green solid is bounded between two paraboloids. The top surface (which opens downward from \(z=8\)) is \(z = 8 - x^2 - y^2\text{.}\) The bottom surface (which opens upward from the origin) is \(z = x^2 + y^2\text{.}\) Because of the rotational symmetry, cylindrical coordinates are perfect here.

🔗

In cylindrical coordinates, the top surface is \(z = 8 - r^2\) and the bottom surface is \(z = r^2\text{.}\) To find the projection of the solid onto the \(xy\)-plane, we set the two \(z\)-equations equal to find their intersection curve:

\begin{align*} 8 - r^2 \amp= r^2 \\ 8 \amp= 2r^2 \implies r^2 = 4 \implies r = 2 \end{align*}

So the projection is a disk of radius 2. Our bounds are:

\begin{align*} 0 \amp\leq \theta \leq 2\pi \\ 0 \amp\leq r \leq 2 \\ r^2 \amp\leq z \leq 8 - r^2 \end{align*}

🔗

We set up the iterated integral. Since we are finding volume, the function is \(1\text{.}\) With \(dV = r \, dz \, dr \, d\theta\text{,}\) the integral is:

\begin{gather*} V = \int_0^{2\pi} \int_0^2 \int_{r^2}^{8-r^2} r \, dz \, dr \, d\theta \end{gather*}

🔗

Evaluate the inner integral with respect to \(z\text{:}\)

\begin{align*} \int_{r^2}^{8-r^2} r \, dz \amp= \Big[ rz \Big]_{z=r^2}^{z=8-r^2} \\ \amp= r(8 - r^2) - r(r^2) \\ \amp= 8r - 2r^3 \end{align*}

🔗

Evaluate the middle integral with respect to \(r\text{:}\)

\begin{align*} \int_0^2 (8r - 2r^3) \, dr \amp= \left[ 4r^2 - \frac{1}{2}r^4 \right]_0^2 \\ \amp= \left( 4(4) - \frac{1}{2}(16) \right) - 0 \\ \amp= 16 - 8 = 8 \end{align*}

🔗

Evaluate the outer integral with respect to \(\theta\text{:}\)

\begin{align*} \int_0^{2\pi} 8 \, d\theta \amp= \Big[ 8\theta \Big]_0^{2\pi} = 16\pi \end{align*}

🔗

15.4.49.

Use spherical coordinates to calculate

\begin{equation*} \iiint_\c{W} \sqrt{x^2 + y^2 + z^2}\, dV \end{equation*}

where \(\c{W} = \left\{ (x,y,z) \in \R^3 \mid x^2 + y^2 + z^2 \leq 2z \right\}\text{.}\)

🔗

Solution.

The region \(\c{W}\) is bounded by \(x^2 + y^2 + z^2 \leq 2z\text{.}\) Let’s convert this inequality into spherical coordinates. We know \(x^2 + y^2 + z^2 = \rho^2\) and \(z = \rho\cos\phi\text{.}\)

\begin{align*} \rho^2 \amp \leq 2\rho\cos\phi \\ \rho \amp \leq 2\cos\phi \end{align*}

🔗

Because \(\rho\) represents a positive distance, we must have \(\rho \geq 0\text{,}\) which means \(2\cos\phi \geq 0\text{.}\) This restricts \(\phi\) to the range \(0 \leq \phi \leq \frac{\pi}{2}\text{.}\) There are no restrictions on \(\theta\text{,}\) so it sweeps completely around from \(0\) to \(2\pi\text{.}\) (Geometrically, this region is a sphere of radius 1 sitting exactly on top of the origin, centered at \((0,0,1)\)).

🔗

The integrand is \(\sqrt{x^2 + y^2 + z^2} = \sqrt{\rho^2} = \rho\text{.}\) Including the spherical volume differential \(dV = \rho^2\sin\phi \, d\rho \, d\phi \, d\theta\text{,}\) the integral is:

\begin{align*} \iiint_\c{W} \sqrt{x^2 + y^2 + z^2} \, dV \amp= \int_0^{2\pi} \int_0^{\pi/2} \int_0^{2\cos\phi} (\rho) \rho^2\sin\phi \, d\rho \, d\phi \, d\theta \\ \amp= \int_0^{2\pi} \int_0^{\pi/2} \int_0^{2\cos\phi} \rho^3\sin\phi \, d\rho \, d\phi \, d\theta \end{align*}

🔗

Evaluate the inner integral with respect to \(\rho\text{:}\)

\begin{align*} \int_0^{2\cos\phi} \rho^3\sin\phi \, d\rho \amp= \left[ \frac{1}{4}\rho^4\sin\phi \right]_{\rho=0}^{\rho=2\cos\phi} \\ \amp= \frac{1}{4}(2\cos\phi)^4\sin\phi - 0 = 4\cos^4\phi\sin\phi \end{align*}

🔗

Evaluate the middle integral with respect to \(\phi\text{.}\) Let \(u = \cos\phi\text{,}\) then \(du = -\sin\phi \, d\phi\text{.}\) The bounds change from \(\cos(0)=1\) to \(\cos(\pi/2)=0\text{.}\)

\begin{align*} \int_0^{\pi/2} 4\cos^4\phi\sin\phi \, d\phi \amp= -4 \int_1^0 u^4 \, du = 4 \int_0^1 u^4 \, du \\ \amp= 4 \left[ \frac{1}{5}u^5 \right]_0^1 = \frac{4}{5} \end{align*}

🔗

Evaluate the outer integral with respect to \(\theta\text{:}\)

\begin{align*} \int_0^{2\pi} \frac{4}{5} \, d\theta \amp= \left[ \frac{4}{5}\theta \right]_0^{2\pi} = \frac{8\pi}{5} \end{align*}

🔗

15.4.53.

Calculate the integral of

\begin{equation*} f(x,y,z) = z\lp x^2 + y^2 + z^2 \rp^{-\frac{3}{2}} \end{equation*}

over the part of the ball \(x^2 + y^2 + z^2 \leq 16\) defined by \(z \geq 2\text{.}\)

🔗

Solution.

The region is the part of the ball \(x^2 + y^2 + z^2 \leq 16\) that lies above the plane \(z = 2\text{.}\) In spherical coordinates, the ball translates to \(\rho \leq 4\text{.}\) The plane translates to \(\rho\cos\phi = 2 \implies \rho = \frac{2}{\cos\phi} = 2\sec\phi\text{.}\)

🔗

To find the angle \(\phi\) where these two bounding surfaces intersect, we set them equal:

\begin{align*} 4 \amp= \frac{2}{\cos\phi} \implies \cos\phi = \frac{1}{2} \implies \phi = \frac{\pi}{3} \end{align*}

So, the angle \(\phi\) drops from the \(z\)-axis (\(\phi = 0\)) out to the intersection edge (\(\phi = \pi/3\)). For any fixed angle \(\phi\) in this range, a ray from the origin pierces into the solid through the flat plane (\(\rho = 2\sec\phi\)) and exits through the spherical dome (\(\rho = 4\)). \(\theta\) sweeps the full circle from \(0\) to \(2\pi\text{.}\)

🔗

The integrand is \(z(x^2 + y^2 + z^2)^{-3/2}\text{.}\) Substituting \(z = \rho\cos\phi\) and \(x^2 + y^2 + z^2 = \rho^2\text{,}\) the function becomes:

\begin{gather*} (\rho\cos\phi)(\rho^2)^{-3/2} = (\rho\cos\phi)(\rho^{-3}) = \rho^{-2}\cos\phi \end{gather*}

🔗

Including \(dV = \rho^2\sin\phi \, d\rho \, d\phi \, d\theta\text{,}\) the integral is:

\begin{align*} \int_0^{2\pi} \int_0^{\pi/3} \int_{2\sec\phi}^4 \big(\rho^{-2}\cos\phi\big) \big(\rho^2\sin\phi\big) \, d\rho \, d\phi \, d\theta \amp= \int_0^{2\pi} \int_0^{\pi/3} \int_{2\sec\phi}^4 \cos\phi\sin\phi \, d\rho \, d\phi \, d\theta \end{align*}

🔗

Evaluate the inner integral with respect to \(\rho\text{:}\)

\begin{align*} \int_{2\sec\phi}^4 \cos\phi\sin\phi \, d\rho \amp= \Big[ \rho\cos\phi\sin\phi \Big]_{\rho=2\sec\phi}^{\rho=4} \\ \amp= 4\cos\phi\sin\phi - (2\sec\phi)\cos\phi\sin\phi \\ \amp= 4\cos\phi\sin\phi - 2\left(\frac{1}{\cos\phi}\right)\cos\phi\sin\phi \\ \amp= 4\cos\phi\sin\phi - 2\sin\phi \end{align*}

We can rewrite \(4\cos\phi\sin\phi\) as \(2\sin(2\phi)\) using the double angle identity. This gives \(2\sin(2\phi) - 2\sin\phi\text{.}\)

🔗

Evaluate the middle integral with respect to \(\phi\text{:}\)

\begin{align*} \int_0^{\pi/3} \big(2\sin(2\phi) - 2\sin\phi\big) \, d\phi \amp= \Big[ -\cos(2\phi) + 2\cos\phi \Big]_0^{\pi/3} \\ \amp= \left( -\cos(2\pi/3) + 2\cos(\pi/3) \right) - \left( -\cos(0) + 2\cos(0) \right) \\ \amp= \left( -\left(-\frac{1}{2}\right) + 2\left(\frac{1}{2}\right) \right) - \left( -1 + 2(1) \right) \\ \amp= \left( \frac{1}{2} + 1 \right) - (1) = \frac{1}{2} \end{align*}

🔗

Evaluate the outer integral with respect to \(\theta\text{:}\)

\begin{align*} \int_0^{2\pi} \frac{1}{2} \, d\theta \amp= \left[ \frac{1}{2}\theta \right]_0^{2\pi} = \pi \end{align*}

🔗

Prev Top Next