Chapter 13

Explain the difference between the roles \(r,\) in cylindrical coordinates, and \(\rho\), in spherical coordinates, play in determining the location of a point.

Why is it easy to use "mass" and "weight" interchangeably, even though they are different measures?

An integral can be interpreted as giving the signed area over an interval; a double integral can be interpreted as giving the signed ________ over a region.

When integrating \(f_{x}(x, y)\) with respect to \(x\), the constant of integration \(C\) is really which: \(C(x)\) or \(C(y) ?\) What does this mean?

Why are points on the \(z\) -axis not determined uniquely when using cylindrical and spherical coordinates?

Give an informal interpretation of what \(" \iiint_{D} d V^{\prime \prime}\) means.

Why would one be interested in evaluating a double integral with polar coordinates?

Explain why the following statement is false: "Fubini's Theorem states that \(\int_{a}^{b} \int_{g_{1}(x)}^{g_{2}(x)} f(x, y) d y d x\) \(\int_{a}^{b} \int_{g_{1}(y)}^{g_{2}(y)} f(x, y) d x d y . "\)

Integrating an integral is called ______ ______ .

What surfaces are naturally defined using cylindrical coordinates?

Give two uses of triple integration.

Explain why if \(f(x, y)>0\) over a region \(R,\) then \(\iint_{R} f(x, y) d A>0\).

If an object has a constant density \(\delta\) and a volume \(V\), what is its mass?

Why is it important to know how to set up a double integral to compute surface area, even if the resulting integral is hard to evaluate?

In Exercises \(3-10\), a function \(f(x, y)\) is given and a region \(R\) of the \(x-y\) plane is described. Set up and evaluate \(\iint_{R} f(x, y) d A\) using polar coordinates. $$ \begin{aligned} &f(x, y)=4 x+4 y ; R \text { is the region enclosed by the circle }\\\ &x^{2}+y^{2}=4 \end{aligned} $$

One understanding of an iterated integral is that \(\int_{a}^{b} \int_{g_{1}(x)}^{g_{2}(x)} d y d x\) gives the ______ of a plane region.

If \(\iint_{R} f(x, y) d A=\iint_{R} g(x, y) d A,\) does this imply \(f(x, y)=\) \(g(x, y) ?\)

Points are given in either the rectangular, cylindrical or spherical coordinate systems. Find the coordinates of the points in the other systems. (a) Points in rectangular coordinates: (2,2,1) and \((-\sqrt{3}, 1,0)\) (b) Points in cylindrical coordinates: \((2, \pi / 4,2)\) and \((3,3 \pi / 2,-4)\) (c) Points in spherical coordinates: \((2, \pi / 4, \pi / 4)\) and (1,0,0)

Two surfaces \(f_{1}(x, y)\) and \(f_{2}(x, y)\) and a region \(R\) in the \(x, y\) plane are given. Set up and evaluate the double integral that finds the volume between these surfaces over \(R\). \(f_{1}(x, y)=8-x^{2}-y^{2}, f_{2}(x, y)=2 x+y ;\) \(R\) is the square with corners (-1,-1) and (1,1) .

Why do \(z=f(x, y)\) and \(z=g(x, y)=f(x, y)+h,\) for some real number \(h\), have the same surface area over a region \(R ?\)

Evaluate the integral and subsequent iterated integral. (a) \(\int_{2}^{3}\left(6 x^{2}+4 x y-3 y^{2}\right) d y\) (b) \(\int_{-3}^{-2} \int_{2}^{5}\left(6 x^{2}+4 x y-3 y^{2}\right) d y d x\)

(a) Evaluate the given iterated integral, and (b) rewrite the integral using the other order of integration. $$ \int_{1}^{2} \int_{-1}^{1}\left(\frac{x}{y}+3\right) d x d y $$

Points are given in either the rectangular, cylindrical or spherical coordinate systems. Find the coordinates of the points in the other systems. (a) Points in rectangular coordinates: (0,1,1) and (-1,0,1) (b) Points in cylindrical coordinates: \((0, \pi, 1)\) and \((2,4 \pi / 3,0)\) (c) Points in spherical coordinates: \((2, \pi / 6, \pi / 2)\) and \((3, \pi, \pi)\)

Two surfaces \(f_{1}(x, y)\) and \(f_{2}(x, y)\) and a region \(R\) in the \(x, y\) plane are given. Set up and evaluate the double integral that finds the volume between these surfaces over \(R\). \(f_{1}(x, y)=x^{2}+y^{2}, f_{2}(x, y)=-x^{2}-y^{2}\); \(R\) is the square with corners (0,0) and (2,3) .

Let \(z=f(x, y)\) and \(z=g(x, y)=2 f(x, y) .\) Why is the surface area of \(g\) over a region \(R\) not twice the surface area of \(f\) over \(R ?\)

In Exercises \(3-10\), a function \(f(x, y)\) is given and a region \(R\) of the \(x-y\) plane is described. Set up and evaluate \(\iint_{R} f(x, y) d A\) using polar coordinates. \(f(x, y)=4 ; R\) is the region enclosed by the petal of the rose curve \(r=\sin (2 \theta)\) in the first quadrant.

Describe a situation where the center of mass of a lamina does not lie within the region of the lamina itself.

Evaluate the integral and subsequent iterated integral. (a) \(\int_{0}^{\pi}(2 x \cos y+\sin x) d x\) (b) \(\int_{0}^{\pi / 2} \int_{0}^{\pi}(2 x \cos y+\sin x) d x d y\)

(a) Evaluate the given iterated integral, and (b) rewrite the integral using the other order of integration. $$ \int_{-\pi / 2}^{\pi / 2} \int_{0}^{\pi}(\sin x \cos y) d x d y $$

Describe the curve, surface or region in space determined by the given bounds. Bounds in cylindrical coordinates: (a) \(r=1, \quad 0 \leq \theta \leq 2 \pi, \quad 0 \leq z \leq 1\) (b) \(1 \leq r \leq 2, \quad 0 \leq \theta \leq \pi, \quad 0 \leq z \leq 1\) Bounds in spherical coordinates: (c) \(\rho=3, \quad 0 \leq \theta \leq 2 \pi, \quad 0 \leq \varphi \leq \pi / 2\) (d) \(2 \leq \rho \leq 3, \quad 0 \leq \theta \leq 2 \pi, \quad 0 \leq \varphi \leq \pi\)

Two surfaces \(f_{1}(x, y)\) and \(f_{2}(x, y)\) and a region \(R\) in the \(x, y\) plane are given. Set up and evaluate the double integral that finds the volume between these surfaces over \(R\). \(f_{1}(x, y)=\sin x \cos y, f_{2}(x, y)=\cos x \sin y+2\); \(R\) is the triangle with corners \((0,0),(\pi, 0)\) and \((\pi, \pi)\).

Set up the iterated integral that computes the surface area of the given surface over the region \(R .\) \(f(x, y)=\sin x \cos y ; \quad R\) is the rectangle with bounds \(0 \leq\) \(x \leq 2 \pi, \quad 0 \leq y \leq 2 \pi\).

In Exercises \(7-10,\) point masses are given along a line or in the plane. Find the center of \(\operatorname{mass} \bar{x}\) or \((\bar{x}, \bar{y}),\) as appropriate. (All masses are in grams and distances are in cm.) $$ m_{1}=4 \text { at } x=1 ; \quad m_{2}=3 \text { at } x=3 ; \quad m_{3}=5 \text { at } x=10 $$

In Exercises \(3-10\), a function \(f(x, y)\) is given and a region \(R\) of the \(x-y\) plane is described. Set up and evaluate \(\iint_{R} f(x, y) d A\) using polar coordinates. $$ \begin{aligned} &f(x, y)=\ln \left(x^{2}+y^{2}\right) ; R \text { is the annulus enclosed by the cir- }\\\ &\text { cles } x^{2}+y^{2}=1 \text { and } x^{2}+y^{2}=4 \end{aligned} $$

Evaluate the integral and subsequent iterated integral. (a) \(\int_{1}^{x}\left(x^{2} y-y+2\right) d y\) (b) \(\int_{0}^{2} \int_{1}^{x}\left(x^{2} y-y+2\right) d y d x\)

(a) Evaluate the given iterated integral, and (b) rewrite the integral using the other order of integration. $$ \int_{0}^{4} \int_{0}^{-x / 2+2}\left(3 x^{2}-y+2\right) d y d x $$

Describe the curve, surface or region in space determined by the given bounds. Bounds in cylindrical coordinates: (a) \(1 \leq r \leq 2, \quad \theta=\pi / 2, \quad 0 \leq z \leq 1\) (b) \(r=2, \quad 0 \leq \theta \leq 2 \pi, \quad z=5\) Bounds in spherical coordinates: (c) \(0 \leq \rho \leq 2, \quad 0 \leq \theta \leq \pi, \quad \varphi=\pi / 4\) (d) \(\rho=2, \quad 0 \leq \theta \leq 2 \pi, \quad \varphi=\pi / 6\)

Two surfaces \(f_{1}(x, y)\) and \(f_{2}(x, y)\) and a region \(R\) in the \(x, y\) plane are given. Set up and evaluate the double integral that finds the volume between these surfaces over \(R\). \(f_{1}(x, y)=2 x^{2}+2 y^{2}+3, f_{2}(x, y)=6-x^{2}-y^{2}\); \(R\) is the disk bounded by \(x^{2}+y^{2}=1\)

Set up the iterated integral that computes the surface area of the given surface over the region \(R .\) \(f(x, y)=\frac{1}{x^{2}+y^{2}+1} ; \quad R\) is bounded by the circle \(x^{2}+\) \(y^{2}=9\).

Point masses are given along a line or in the plane. Find the center of \(\operatorname{mass} \bar{x}\) or \((\bar{x}, \bar{y}),\) as appropriate. (All masses are in grams and distances are in cm.) $$ \begin{array}{l} m_{1}=2 \text { at } x=-3 ; \quad m_{2}=2 \text { at } x=-1; \\ m_{3}=3 \text { at } x=0 ; \quad m_{4}=3 \text { at } x=7 \end{array} $$

In Exercises \(3-10\), a function \(f(x, y)\) is given and a region \(R\) of the \(x-y\) plane is described. Set up and evaluate \(\iint_{R} f(x, y) d A\) using polar coordinates. $$ \begin{aligned} &f(x, y)=1-x^{2}-y^{2} ; R \text { is the region enclosed by the circle }\\\ &x^{2}+y^{2}=1 \end{aligned} $$

Evaluate the integral and subsequent iterated integral. (a) \(\int_{y}^{y^{2}}(x-y) d x\) (b) \(\int_{-1}^{1} \int_{y}^{y^{2}}(x-y) d x d y\)

(a) Evaluate the given iterated integral, and (b) rewrite the integral using the other order of integration. $$ \int_{1}^{3} \int_{y}^{3}\left(x^{2} y-x y^{2}\right) d x d y $$

Set up the triple integrals that give the volume of \(D\) in all 6 orders of integration, and find the volume of \(D\) by evaluating the indicated triple integral. \(D\) is bounded by the coordinate planes and \(z=2-2 x / 3-2 y\).

Set up the iterated integral that computes the surface area of the given surface over the region \(R .\) \(f(x, y)=x^{2}-y^{2} ; \quad R\) is the rectangle with opposite corners (-1,-1) and (1,1).

Point masses are given along a line or in the plane. Find the center of \(\operatorname{mass} \bar{x}\) or \((\bar{x}, \bar{y}),\) as appropriate. (All masses are in grams and distances are in cm.) $$ \begin{array}{l} m_{1}=2 \text { at }(-2,-2) ; \quad m_{2}=2 \text { at }(2,-2) ; \\ m_{3}=20 \text { at }(0,4) \end{array} $$

In Exercises \(3-10\), a function \(f(x, y)\) is given and a region \(R\) of the \(x-y\) plane is described. Set up and evaluate \(\iint_{R} f(x, y) d A\) using polar coordinates. \(f(x, y)=x^{2}-y^{2} ; R\) is the region enclosed by the circle \(x^{2}+y^{2}=36\) in the first and fourth quadrants.

Evaluate the integral and subsequent iterated integral. (a) \(\int_{0}^{y}(\cos x \sin y) d x\) (b) \(\int_{0}^{\pi} \int_{0}^{y}(\cos x \sin y) d x d y\)

(a) Evaluate the given iterated integral, and (b) rewrite the integral using the other order of integration. $$ \int_{0}^{1} \int_{-\sqrt{1-y}}^{\sqrt{1-y}}(x+y+2) d x d y $$

Set up the triple integrals that give the volume of \(D\) in all 6 orders of integration, and find the volume of \(D\) by evaluating the indicated triple integral. \(D\) is bounded by the planes \(y=0, y=2, x=1, z=0\) and \(z=(3-x) / 2\) Evaluate the triple integral with order \(d x d y d z\).