Presentations
Pacing Tracker
- The quizzes have included questions for 22 objectives. How many have you passed? What are you plans to master those that you haven't mastered yet?
- We've finished units 1 through 4. Have you started your self-directed learning project for each unit?
- The 6th project can be over any topic from the entire semester. Feel free to get started on this one as soon as you have an idea.
- Remember you can submit only one SDL project per week. Plan ahead and don't let yourself get behind.
Brain Gains
- Set up an iterated integral that computes the area of the region $R$ that is inside a circle of radius $a$ centered at the origin, and left of the $y$-axis.
Solution
$$A = \int_{\pi/2}^{3\pi/2}\int_0^a r\,dr\,d\theta = \int_{-a}^a\int_{-\sqrt{a^2-y^2}}^0\,dx\,dy = \int_{-a}^0\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dy\,dx$$
- Suppose the region $R$ above describes a laminar object with density $\delta(x,y) = 5+x^2+y^2$. Set up a double integral formula to compute the mass of the object.
Solution
$$m = \int_{\pi/2}^{3\pi/2}\int_0^a (5+r^2)\, r\,dr\,d\theta = \int_{-a}^a\int_{-\sqrt{a^2-y^2}}^05+x^2+y^2\,dx\,dy = \int_{-a}^0\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}5+x^2+y^2\,dy\,dx$$
- Consider two vectors $\vec u = \left< a,b,0 \right>$ and $\vec v = \left< c,d,0 \right>$ in the $xy$-plane. They form a parallelogram in the $xy$ plane.
- Geometrically explain why the cross product either points directly up, or directly down (so is parallel to the $z$-axis).
- Give the magnitude of the cross product of $\vec u$ and $\vec v$.
- Use the cross product formula $\vec u \times \vec v = \left<u_2 v_3 - u_3 v_2, u_3 v_1 - u_1 v_3, u_1 v_2 - u_2 v_1\right>$ to compute the cross product.
Solution
- The cross product gives a vector that is orthogonal to both. The $z$-axis meets any vector in the $xy$-plane at a 90 degree angle.
- Because the cross product has a magnitude equal to the area of the parallelogram formed, we can use $A = |ad-bc|$.
- Using the formula gives $$\vec u \times \vec v = \left<b 0 - 0 d, 0 c - a 0, a d - b c\right>= \left<0, 0, a d - b c\right>.$$ Note that the length of this vector is $|\vec u \times \vec v| = \sqrt{(ad-bc)^2}=|ad-bc|.$
Group problems
- Consider $\int_{0}^{4}\int_{x}^{4}e^{y^2}dydx$.
- Draw the region described by the bounds.
- Swap the order of the bounds on the integral (use $dxdy$ instead of $dydx$).
- Compute the integral.
- Draw the region described the bounds of each integral. (Use the Mathematica notebook Integration.nb to check your work.)
- $\ds\int_{0}^{3}\int_{0}^{9-x^2}\int_{0}^{5}dzdydx$
- $\ds\int_{0}^{3}\int_{0}^{9-x^2}\int_{0}^{3-x}dzdydx$
- $\ds\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{\sqrt{1-x^2}}dydxdz$
- $\ds\int_{0}^{\pi}\int_{0}^{3}\int_{0}^{5}rdzdrd\theta$
- $\ds\int_{-1}^{1}\int_{0}^{1-y^2}\int_{0}^{x}dzdxdy$
- Set up an integral formula to compute each of the following:
- The mass of a disc that lies inside the circle $x^2+y^2=9$ and has density function given by $\delta = x+10$
- The $x$-coordinate of the center of mass (so $\bar x$) of the disc above.
- The $z$-coordinate of the center-of-mass (so $\bar z$) of the solid object in the first octant (all variables positive) that lies under the plane $2x+3y+6z=6$ (so $\frac{x}{3}+\frac{y}{2}+\frac{z}{1}=1$).
- The $y$-coordinate of the center-of-mass (so $\bar y$) of the same object.
- A wire lies along the curve $C$ parametrized by $\vec r(t) = (t^2+1, 3t, t^3)$ for $-1\leq t\leq 2$.
- Compute $ds$. (Remember - a little distance equals the product of the speed and a little time.)
- Set up an integral to find $\bar x$, then $\bar y$, then $\bar z$, for the centroid of $C$.
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