Worksheet #1. A little review.

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Brice Kelley
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1 Worksheet #1. A little review. I. Set up BUT DO NOT EVALUATE definite integrals for each of the following. 1. The area between the curves = 1 and = 3. Solution. The first thing we should ask ourselves is Where is this area?. Let s plot the curves. = 1 is a hperbola and = 3 is a line with slope that goes through the point (, 3). What is the relative position of the line and the hperbola? Three possibilities come to mind: Hopefull we are in the third situation (otherwise does it even make sense to talk of the area between the two curves?). In order to check whether we are, we have to compute where the two graphs intersect. This amounts to solving the equation 1 = 3 : = 1 = 3 1 = 3 (Multipl b, which will never equal ) = 3 ± 3 4 = 1 or 1/ (Quadratic formula) So indeed there s two intersection points, which have -coordinates 1/ and 1, respectivel. This means that we are in the third situation from the pictures above. The area between the curves is the integral of the function on top minus the curve on the bottom, and the endpoints must be 1/ and 1, so our answer is Area = 1 1/ 3 1 d. The volume of the solid generated b revolving the region R, bounded b = sin, =, and = π, about the -ais. 1

Solution. First of all, grab a piece of paper, or even better, our canvas and oil paints, and make sure ou understand wh ever picture looks like it does. Let s start b drawing the region R. 1.5.

2 Solution. First of all, grab a piece of paper, or even better, our canvas and oil paints, and make sure ou understand wh ever picture looks like it does. Let s start b drawing the region R The solid made b revolving around the -ais looks sort of like Have ou ever tried googling online surface plot? I have! Recall the formula for the volume of a solid of revolution that has radius f() at height, and is bounded between heights a and b, is b a f() d In this case, our bounds are and π/, so the integral that computes the volume of the solid is π/ sin ()d

3 3. The volume of the solid generated b revolving the region R, bounded b = sin, =, and = π, about the line = π. Solution. The region R is the same as before. However, if we rotate it around the line = π, we get something like a donut, whose outside walls look like Now our solid has a hole in the middle, so recall the formula from calc 1, that sas that the volume is b a r out r ind Notice that the ais of revolution is vertical, so we have to integrate with respect to! The limits will thus be and 1. We have to use two distances: 1 r out.5 r in

4 r in is ver eas: it s just π/ for ever. What about r out?. Look at the coordinate of the two points: 1 (, sin()) (π, sin() (Think about it: wh are these the coordinates?) The distance between these points is the difference between their -coordinates, that is, r out = π. However, we need to integrate with respect to, so we need to give the answer in terms of, the height. How do we do this? Well, = sin(), so = arcsin(). Therefore, Putting everthing together, b a r out = π = π arcsin() r out r ind = π 1 ( π ) (π arcsin()) d II. Find the absolute maimum and minimum values of f() = on the interval [, 1]. Solution. We just need to evaluate the function at all the critical points in [, 1] and at the endpoints. To find the critical values, we look for the points where f () = : = f () = = ± = 3 or 1 1 is not in the interval, so we just have to evaluate the function at, 3 f() = f(/3) = 44 7 f(1) = 1 So the maimum value is and the minimum value is and 1.

5 III. Consider the points A = (1, 1, 1), B = (1,, ) and C = (3,, 3) in space. 1. Find an equation for the plane containing A, B and C. Solution. Recall that the equation of a plane with normal vector n going through a point a is n ( a) =. We alread have point, so we need a normal vector. Recall that the cross product of two vectors is normal to both of them. Do we have two vectors contained in the plane? Yes! The vectors AB and AC, for eample, are two vectors contained in the plane. Thus we can take Now we just have to compute things: n = AB AC AB = B A = (, 1, 1) AC = C B = (, 1, ) i j k n = AB AC = det 1 1 = 3 i + j k = ( 3,, ) 1 Since it s ver eas to make a mistake taking a cross product, I m going to check that n is perpendicular to AB and AC, as it should be. The are perpendicular if their dot product is : AB n = (, 1, 1) ( 3,, ) = = AC n = (, 1, ) ( 3,, ) = = 5

6 I didn t make a mistake computing n, so the equation for the plane is (one of man possible answers) = n ( a) = ( 3,, ) ((,, z) (1, 1, 1)) If we simplif it, it becomes the equation 3 + z = 1. What is the area of the triangle formed b these points? Solution. Recall that the area of a triangle ABC equals 1 AB computed in the previous part its length: AC. Luckil, we AB AC = ( 3,, ), so we onl need to compute Area(ABC) = 1 AB AC = 1 ( 3,, ) = =

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