Saturday

April 19, 2014

April 19, 2014

Number of results: 75,848

**Calculus-Area between curves **

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 2y=4*sqrt(x) , y=5 and 2y+4x=8 please help! i've been trying this problem the last couple days, even asked a TA for help, but i can't arrive at...
*Thursday, November 21, 2013 at 10:16pm by Chrissy*

**calculus**

consider the area enclosed between the curves f(x)=x^2 and g(x)=4x what is the volume obtained by revolving the area between these two curves around the line y=20 ? please help, i don't know what to do
*Tuesday, February 2, 2010 at 5:02pm by amy*

**calculus**

consider the area enclosed between the curves f (x) = x2 and g (x) = 4x what is the volume obtained by revolving the area between these two curves around the line y = 20 ? please help, i don't know what to do
*Tuesday, February 2, 2010 at 1:07pm by amy*

**Calculus (Area Between Curves)**

Find the area of the region bounded by the curves y^2=x, y-4=x, y=-2 and y=1 (Hint: You'll definitely have to sketch this one on paper first.) You get: a.) 27/2 b.) 22/3 c.) 33/2 d.) 34/3 e.) 14
*Wednesday, February 29, 2012 at 2:39pm by Mishaka*

**Calculus Area between curves**

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 3y+x=3 , y^2-x=1
*Thursday, November 21, 2013 at 10:22pm by Amy*

**Calculus (Area Between Curves)**

Find the area of the region IN THE FIRST QUADRANT (upper right quadrant) bounded by the curves y=sin(x)cos(x)^2, y=2xcos(x^2) and y=4-4x. You get: a.)1.8467 b.) 0.16165 c.) 0.36974 d.) 1.7281 e.) 0.37859
*Wednesday, February 29, 2012 at 4:43pm by Mishaka*

**calculus**

The curves intersect at x = 0 and x = 1. The region bounded between those curves has y-separation of 6(x-x^6). For the total enclosed area, integrate that function times dx from x=0 to x=1.
*Friday, January 30, 2009 at 8:33pm by drwls*

**math**

consider the area enclosed between the curves f (x) = x2 and g (x) = 4x what is the volume obtained by revolving the area between these two curves around the line y = 20?
*Tuesday, February 2, 2010 at 11:57pm by amy*

**Calculus (Area Between Curves)**

Find the area of the region IN THE FIRST QUADRANT (upper right quadrant) bounded by the curves y=sin(x)cos(x)^2, y=2xcos(x^2) and y=4-4x. You get: a.)1.8467 b.) 0.16165 c.) 0.36974 d.) 1.7281 e.) 0.37859 Based on my calculations, I would say that the answer is e.) 0.37859. I ...
*Wednesday, February 29, 2012 at 4:21pm by Mishaka*

**Calculus (Area Between Curves)**

You're welcome! :)
*Wednesday, February 29, 2012 at 4:21pm by MathMate*

**calculus area between three curves**

also help with Find c>0 such that the area of the region enclosed by the parabolas y=x^2-c^2 and y=c^2-x^2 is 430.
*Monday, December 2, 2013 at 2:12pm by jaime*

**Calculus (Area Between Curves)**

Find the area of the region bounded by the curves y=x^(-1/2), y=x^(-2), y=1 and y=3. You get: a.) 1/2(sqrt(3)) + 4/3 b.) 2(sqrt(3)) - 8/3 c.) 1/2(sqrt(3) - 32/3 d.) 2(sqrt(3)) - 32/3 e.) 8/3 - 2(sqrt(3))
*Wednesday, February 29, 2012 at 3:44pm by Mishaka*

**maths**

Note that the two curves intersect at x = 2 and x = 5. You only need to evaluate the area intercepted between those points. It is the area between an inverted parabola and a 45 degree sloped straight line. The area is easily calculated with calculus or numerical integration. ...
*Monday, April 11, 2011 at 12:14am by drwls*

**Calculus (Area Between Curves)**

33/2 is correct.
*Wednesday, February 29, 2012 at 2:39pm by Steve*

**calculus**

Calculate the area between curves y=e^x and y=-e^x on the interval [0,1]
*Thursday, December 13, 2012 at 4:50pm by Kieran*

**Calculus Area between curves**

Consider the area between the graphs x+6y=8 and x+8=y2. This area can be computed in two different ways using integrals First of all it can be computed as a sum of two integrals where a= , b=, c= and f(x)= g(x)= I found a, but not b or c. I can't seem to figure out f(x) and g(...
*Sunday, November 24, 2013 at 10:20pm by Kelly*

**Calculus**

1. Find the area of the region between the curves y=sin(x pi/2) and y=x. 2. Find the area of the region between the curves y=sin(x), y=sin(2x), x=0, and x=pi/2.
*Thursday, January 24, 2013 at 7:46pm by Anonymous*

**Calculus**

1. Find the area of the region between the curves y=sin(x pi/2) and y=x. 2. Find the area of the region between the curves y=sin(x), y=sin(2x), x=0, and x=pi/2.
*Friday, January 25, 2013 at 2:41am by Anonymous*

**Calculus**

1. Find the area of the region between the curves y=sin(x pi/2) and y=x. 2. Find the area of the region between the curves y=sin(x), y=sin(2x), x=0, and x=pi/2.
*Friday, January 25, 2013 at 2:41am by Anonymous*

**Calculus (Area Between Curves)**

I got the same thing. Working out the other answers, they were either negative or obviously too large of an area for the given bounds, thank you Nade!
*Wednesday, February 29, 2012 at 3:44pm by Mishaka*

**calculus area between three curves**

find the area of the region for 2y=4sqrt(x) , y=3, and 2y+3x=7
*Monday, December 2, 2013 at 2:12pm by jaime*

**Calculus (Area Between Curves)**

Thank you, I was just really unsure of my answer!
*Wednesday, February 29, 2012 at 4:21pm by Mishaka*

**Calculus-Area between curves**

Thanks I figured it out my answer was 9, and it was correct
*Thursday, November 21, 2013 at 10:16pm by Chrissy*

**Calculus **

You will have to sketch the three curves yourself. The curves y = (3/2)sqrtx and y = (5/2) -x intersect at (1,3/2) The enclosed area starts at that point and expands upward to the line y = 3, bounded by the two other curves along the way. Your best bet is to integrate (x2 - x1...
*Wednesday, August 17, 2011 at 1:49am by drwls*

**Calculus (Area Between Curves)**

Since y^2 = x is not even defined left of the y-axis, the region is x<3: a triangle bounded by y=x+4, x=-3, y=-2 area = 3*3/2 = 9/2 -3<x<0: a rectangle, area 3*3=9 0<=x<=1: a curved triangular area area = Int(1-√x)dx [0,1] = (x - 2/3 x√x)[0,1] = (1...
*Wednesday, February 29, 2012 at 2:39pm by Steve*

**Calculus**

Find the area between the two curves f(x)= ln(x+1)+2 g(x)= x^2-8x+12
*Tuesday, December 23, 2008 at 5:38pm by mary*

**calculus 1**

find the total area between the curves ( y= 4-x ^2 ) and the x-axis over the intervals [0,4]
*Friday, November 19, 2010 at 3:12pm by martin*

**Calculus**

find the area between the two curves y=1/2x and y=x square root of 1-x^2
*Tuesday, September 11, 2012 at 1:00pm by Lauren*

**Calculus (Area Between Curves)**

The definite integral of y^-.5-y^-2 evaluated from 1 to 3. I forgot 2sqrt(3)-8/3 choice b
*Wednesday, February 29, 2012 at 3:44pm by Nade*

**Calculus (Area Between Curves)**

That's the correct answer. If you integrated between the end-points 0, 0.69275, 0.92811, you should have got the areas 0.28024 and 0.09835 respectively, which add up to 0.37859.
*Wednesday, February 29, 2012 at 4:21pm by MathMate*

**CALCULUS ONE!**

The graph of sinx and cosx intersect once between 0 and pi/2. What is the angle between the two curves at the point where they intersect? (You need to think about how the angle between two curves should be defined).
*Tuesday, October 18, 2011 at 12:03am by Sara*

**Calculus (Area Between Curves)**

find the definite integral y^2 - y + 4 evaluated from -2 to 1. I go 16.5 sq units Choice c
*Wednesday, February 29, 2012 at 2:39pm by Nade*

**Calculus**

Find the area between the two curves y1= x^2 - 4x + 5 and y2 = 2x - 3, finding the points of intersection algebraically.
*Monday, May 27, 2013 at 3:39am by Davey*

**calculus**

I think there's a typo here. The curves intersect at (0,0) and (4,2). Going out to x=9 is bogus, unless you want to add up two areas. Think of the area as a lot of small rectangle, each of width dx and height the distance between the curves. a = ∫[0,4] √x - x/2 dx ...
*Sunday, September 8, 2013 at 2:23pm by Steve*

**Calculus Check my answer **

Find the area between the curves y= x^2/2 +2 and y = –x – 3 on the interval –4 ≤ x ≤ 4. I got 40/3 is that correct?
*Tuesday, March 11, 2014 at 9:43pm by Ryan*

**2nd question - calculus area between three curves**

nice symmetry, should be easy to see that the x-intercepts are (c,0) and (-c,0) for both so just double the area from 0 to c area = 2∫(c^2 - x^2 - (x^2 - c^2)) dx from 0 to c = 2∫(2c^2 - 2x^2) dx from 0 to c = 2[2c^2 x - (2/3)x^3] from 0 to c = 2(c^3 - (2/3)c^3] = ...
*Monday, December 2, 2013 at 2:12pm by Reiny*

**calculus wxmaxima**

The two curves intersect at -0.4425 and 1.8018. By integrating (numerically) f(x)=cos(x^2 +(100493/100000)) - (1+x-x^2) between the two roots, you would get the area as 1.99326. The perimeter can be obtained by integrating the curve length of the separate curves between the ...
*Wednesday, February 16, 2011 at 6:59pm by MathMate*

**Calculus kinda with the area of two curves**

the two curves intersect where x^2-c^2 = c^2-x^2 that is, where x=±c The area is thus ∫[-c,c] (c^2-x^2)-(x^2-c^2) dx = 4∫[0,c] c^2-x^2 dx = 4(c^2 x - 1/3 x^3)[0,c] = 4(c^3 - 1/3 c^2) = 8/3 c^3 8/3 c^3 = 270 8c^3 = 30*3^3 c = 3/2 ∛30
*Monday, December 2, 2013 at 2:42pm by Steve*

**Calc**

First find the intersections between the two curves, they are the limits of integration. They can be found readily as x=±2. Next, we check that the curves do not cross the axis y=-1 between the limits of integration. Otherwise the limits of integration must change. Now ...
*Thursday, December 15, 2011 at 7:15pm by MathMate*

**Calculus (Area Between Curves)**

Thank you Nade for making this correction! I went back and graphed the functions with pencil to see if I could figure a rough estimate for the area. I found that these function make an odd trapezoid shape. The area of the solid trapezoid portion of the graph is 15 and then ...
*Wednesday, February 29, 2012 at 2:39pm by Mishaka*

**Calculus - please help!**

Suppose that 0 < c < ¥ð/2. For what value of c is the area of the region enclosed by the curves y = cos x, y = cos(x - c), and x = 0 equal to the area of the region enclosed by the curves y = cos(x - c), x = ¥ð, and y = 0? i have no idea how to solve this question.
*Thursday, November 25, 2010 at 10:12pm by K*

**maths**

determine the co-ordinate of the poin of intersection of the curves y=x*x and y*y=8x. sketch the two curves and find the area enclosed by the two curves.
*Monday, February 7, 2011 at 1:19am by patrick*

**calculus**

the curves intersect at (1,2) and (3,0) A = ∫[1,3] (3-x) - (x^2-5x+6) dx = -x^3/3 + 2x^2 - 3x [1,3] = 4/3 the parabola dips below the x-axis on the interval (2,3), so we want to rotate between the curves on [1,2] and under the line on [2,3]: V = ∫[1,2] pi (3-x)^2...
*Tuesday, December 11, 2012 at 12:54pm by Steve*

**Calc 2**

I don't think the equations enclose any area. But their graphs do. Gotta be precise when talking and doing math! The curves intersect at y=±4, so the area between the curves is just ∫[-4,4] (16-y^2)-(y^2-16) dy which, taking advantage of the symmetry is 4∫[0,4] 16-...
*Saturday, December 7, 2013 at 5:44pm by Steve*

**calculus**

The area is between the function y=x^5 and y=81x, between the limits x=0 and x=3. The functions are all in the first quadrant (no values negative), and the two curves y=81x and y=x^5 intersect at x=3, the upper integration limit, which is a "lucky coincidence". The area ...
*Sunday, November 28, 2010 at 12:43pm by MathMate*

**Calculus**

Find the area between the following curves y1= x2−4x+3 y2=−x2+2x+3
*Monday, December 17, 2012 at 9:57am by Corey*

**Calculus **

Find the area between the curves y=(x^3)-10(x^2)+24x and y=(-x^3)+10(x^2)-24x
*Friday, October 22, 2010 at 8:37pm by Hannah*

**Calculus**

Your intersection points are correct. Between those points, the y = x+4 curve is above the y = x^2 -2x curve. The area between the curves is the integral of x + 4 dx from x = -1 to 4, MINUS the integral of x^2 -2x between the same two x values. That equals x^2/2 + 4x @ x = 4...
*Monday, May 9, 2011 at 10:00pm by drwls*

**calculus**

Consider the curves y = x^2and y = mx, where m is some positive constant. No matter what positive constant m is, the two curves enclose a region in the first quadrant.Without using a calculator, find the positive constant m such that the area of the region bounded by the ...
*Sunday, March 4, 2012 at 2:31pm by John*

**Calculus kinda with the area of two curves**

II don't even know where to start with this can anyone help?!? Find c>0 such that the area of the region enclosed by the parabolas y=x^2-c^2 and y=c^2-x^2 is 270.
*Monday, December 2, 2013 at 2:42pm by Roni*

**Calculus Area between curves**

Evaluate the definite integral: sqrt(8-2x) lower limit=-7 upper limit=0 I got -(1/3)(8-2x)^(3/2) and it was wrong. Please Help! Thanks in advance!
*Friday, November 22, 2013 at 8:05pm by Kelly*

**Calculus**

Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. Between y = −x and y = x/2 for x in [0, 6]
*Thursday, April 12, 2012 at 10:32pm by Alex*

**calculus**

What is the area under each of the given curves? y=e^x/2 ; x=0 x=1
*Thursday, December 1, 2011 at 11:54pm by bob*

**Calculus**

Find the total area bounded by the curves y = x^3 and y = x^5 .
*Sunday, December 4, 2011 at 4:16pm by Beth*

**CALCULUS HELP PLEASE!!**

The question is done correctly. When you differentiate -x^-1 you get x^-2 and all the arithmetic was done correctly. If this an "area between curves" type, I have seen this before. The function f(x) = 1/x^2 is discontinuous at x=0 so even though calculation-wise we come up ...
*Wednesday, May 11, 2011 at 11:12pm by Reiny*

**Calculus - please help!**

yes, ¥ðstands for pi. The quesiton that i wrote above is all information that are given. There is no other limit on this integration. I will wrote the question once again. Please help. Thank You. Suppose that 0 < c < pi/2. For what value of c is the area of the region ...
*Thursday, November 25, 2010 at 10:12pm by K*

**Calculus**

Find the area of the enclosed by the y-axis and the curves y=x^2 and y=(x^2+x+1)*e^(-x). ...I'm supposed to use tabular method to find the area. But I'm not sure where to start with this question. I drew the graph for y=x^2, but that's about it. Help! thanks
*Saturday, October 30, 2010 at 1:49pm by Anonymous*

**math/calc**

area = ∫ x^2 dx from x = 1 to 2 = [(1/3)x^3\ from 1 to 2 = (1/3)(8) - (1/3)(1) = 7/3 google "area under curves" and pick a suitable page that you understand. You will have to know basic Calculus, and basic integration.
*Wednesday, July 18, 2012 at 11:18am by Reiny*

**Calculus - Center of Mass**

Find the exact coordinates of the centroid given the curves: y = 1/x, y = 0, x = 1, x = 2. X = 1/Area*Integral from a to b: x*f(x)dx Y = 1/Area*Integral from a to b: [(1/2)*(f(x))^2]dx How do I find the area for this? Once I know that, is this the correct set up? X = 1/Area*...
*Tuesday, June 26, 2007 at 12:21am by COFFEE*

**calculus**

what is the area of the region bounded by the curves y=x^2 , y=8-x^2 , 4x-y+12=0
*Wednesday, December 1, 2010 at 6:38pm by nicko*

**integral calculus**

FIND THE AREA OF THE REGION BOUNDED BY THE CURVES Y= X^2 + 4X + 3 AND Y= x-1.
*Monday, January 9, 2012 at 11:30pm by Anonymous*

**calculus**

Find the area bounded by {y=x2−4 y=4−x2 • sketch the region described • determine any intersection point(s) for the curves (show work!!) • write out the integral(s) that will calculate the area • determine the area (may use a calculator)
*Thursday, October 3, 2013 at 6:40pm by Will*

**Calculus**

Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. (Round your answer to three decimal places.) Between y = e^x and y = x for x in [1, 2]
*Wednesday, December 11, 2013 at 6:58pm by tiffany*

**calculus**

find the area of the regions enclosed by the lines and curves. x+y^2=0 and x+3y^2=2 i know the answer is 8/3, but i don't know how to get it.
*Thursday, April 29, 2010 at 9:18pm by shelby*

**calculus**

find the area of one of the regions bounded by the curves y=sin x and y=cos x
*Sunday, March 27, 2011 at 6:53pm by monday*

**Calculus**

Find the area of the region enclosed by the given curves: 4x+y^2=9, x=2y
*Thursday, April 19, 2012 at 10:17am by Michael*

**calculus**

Find the area of region bounded by the curves y=sin(pi/2*x)and y=x^2-2x.
*Wednesday, January 30, 2013 at 7:21pm by Liz*

**Calculus Area between curves**

Actually your answer is correct if it's indefinite integral. :) Now that you got the answer, we can now evaluate it at the given bounds: -(1/3)(8-2x)^(3/2) , at x = -7 to 0 = -(1/3)(8-2(0))^(3/2) - [-(1/3)(8-2(-7))^(3/2)] = -(1/3)(8)^(3/2) - [-(1/3)(22)^(3/2)] = (22/3)*(22)^(1...
*Friday, November 22, 2013 at 8:05pm by Jai*

**Calculus**

Find the area of the region enclosed by the given curves: y=e^6x, y=2sin(x), x=0, x=pi/2
*Wednesday, April 18, 2012 at 1:43pm by Michael*

**calculus**

two curves are orthogonal at a point of intersection of their tangents at that point cross at right angles. Show that the curves 2x^2+3y^2=5 and y^2=x^3 are orthogonal at (1,1) and (1,-1). Use parametric mode to draw the curves and to show the tangent lines
*Tuesday, November 15, 2011 at 4:59pm by Anonymous*

**CALC URGENT**

First use algebra to find where the enclosed area is. Then integrate the difference between the functions from one intersection to the other. Drawing a graph will help a lot. It seems to me, after drawing a sketch, that the two curves never intersect, and that the "enclosed ...
*Thursday, January 29, 2009 at 5:22am by drwls*

**calculus**

Find the angle of intersection between the curves f(x) = 2x + 3 and g(x ) = - 3x + 1
*Thursday, August 4, 2011 at 4:39pm by mmk*

**University Maths**

Find the area enclosed between the ffg curves :- y=lnx and y= -2x+3 and the ordinate x=3
*Tuesday, November 8, 2011 at 5:09am by H.A.*

**Math **

Determine the area of the region between the curves: f(x)= 3x^2 + 16x + 28 and g(x) = -2x + 1
*Sunday, May 26, 2013 at 12:57pm by mark *

**math **

Determine the area of the region between the curves: f(x) = -(3)x^3 + (6.25)x^2 + (27.1875)x - (9.328125) and g(x) = (1)x^2 + (3)x -(3)
*Sunday, May 26, 2013 at 7:22pm by mark *

**university mathematics**

find the area enclosed between the following curves:- Y= lnx and y= -2x+3 and the ordinate x=3
*Tuesday, November 1, 2011 at 9:43am by anonymous*

**Calculus**

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 2y = 4(x^(1/2)), y=4, and 2y +2x = 6 I keep getting an area around 21.3 but it is incorrect. Am I close? Thank you!
*Thursday, July 11, 2013 at 9:15pm by Donna*

**Calculus - MathMate Please help**

I asked this question on last thursday and u told me that the bottom limit for the 1st and the right limit for the 2nd area is missing? The question that i posted had no other information and i dont know why you said that the limits are missing. There is no other limits that ...
*Monday, November 29, 2010 at 2:44am by K*

**Math - Calculus III**

Find the double integral of f (x, y) = (x^7)y over the region between the curves y = x^2 and y = x(3 - x).
*Tuesday, November 8, 2011 at 11:55pm by Josh*

**Calculus: Application of Integration**

It is easy to see that the two curves intersect at (0,0) and (1,1) I will integrate with respect to x area = [integral] (x^(1/2) - x^2) dx from 0 to 1 = [(2/3)x^(3/2) - (1/3)x^3] from 0 to 1 = 2/3 - 1/3 - 0 = 1/3
*Tuesday, October 5, 2010 at 11:02pm by Reiny*

**natural logs**

can't use a graphing calculator (not allowed). I'm trying to find the area between curves, if that helps.
*Friday, December 19, 2008 at 4:13pm by Anne*

**calculus **

find the area enclosed by the given curves. y=squareroot of x y=x/2 x=9 * i just graph it but do not know how to set it up the equation or anything.
*Sunday, September 8, 2013 at 2:23pm by can somebody explain how to do it?*

**Calculus**

Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. Between y = x^2 − 4x + 1 and y = −x^2 + 4x − 5 for x in [0, 3] Please don't...
*Monday, December 16, 2013 at 6:13pm by tiffany*

**calculus**

The f(x) curve is above the x-axis from x=0 to the point where 2x = x^2, which is x = 2. The area is INTEGRAL(2x - x^2) dx = x^2 -x^3/3 @x=2 0 to 2 = 4/3. That agrees with what you obtained. You want the area under y = mx to be 2/3. That area will be (1/2)xy. where x and y are...
*Sunday, January 6, 2008 at 8:39am by drwls*

**calculus wxmaxima**

does anyone know how to approximate the are and circumfrence of the region bounded by the given curves y=cos(x^2 +(100493/100000)), y=1+x-X^2 i already did the area but i need help with the circumference
*Wednesday, February 16, 2011 at 8:59pm by jessica*

**calculus**

Determine the exact value for the constant k such that the area of the region bounded by the curves y=x and y=kx^2 is equal to 2/3 sq units. Any help is appreciated.
*Monday, April 18, 2011 at 5:05pm by Jenn*

**Calculus AP**

areas between the curves x=y^2-2, x=y, y=3. int[a,b](f(y)-g(y)) dy i know how to graph. but im have problem with integral.
*Thursday, September 27, 2012 at 9:57pm by Vicky*

**calculus**

Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. Between y = x2 − 4x + 1 and y = −x2 + 4x − 5 for x in [0, 3]
*Sunday, December 15, 2013 at 9:45pm by Leticia*

**calculus**

You do not need Calculus for this question, since this results in a trapezoid. The two parallel sides have lengths of 2 and 18 and the distance between them is 2 Area = 2(2+18)/2 = 20 if you insist on Calculus Area = [integral](8x+2)dx from 0 to 2 = |4x^2 + 2x| from 0 to 2 = ...
*Sunday, December 5, 2010 at 4:33pm by Reiny*

**calculus**

Graph it first. See where the curves meet. If they meet at some angle less than PI/4, then you will have to break up the integration (upper-lower)dx into segments 1->meeting point> otherwise you get a "negative" area which will subtract from your area.
*Saturday, May 1, 2010 at 3:51pm by bobpursley*

**Rosaline have you got it ?**

If you are still stuck, look at the graphs Steve linked you to carefully and see three areas between the curves on the x axis. Those are values of x where your original function is between the limits. The last area on the right is not feasible because the width would be ...
*Tuesday, March 25, 2014 at 8:29pm by Damon*

**math**

Find the area of the region between the curves y = sin x and y = x^2 - x, 0 ≤ x ≤ 2.
*Sunday, January 27, 2013 at 3:10pm by Em*

**Calculus 3**

Given the function f(x,y)=y/(x^2+y^2), give and identify the level curves for k=1, k=1/2, and k=1/4, and draw a contour map showing these level curves. Any help would be immensely appreciated. Thanks
*Sunday, February 26, 2012 at 8:55pm by Anonymous*

**Calculus**

Find the area of the region that lies inside both curves. r = (3)^(1/2) cos (theta), r = sin (theta)
*Wednesday, June 9, 2010 at 10:57pm by Mercedes*

**calculus**

area=1/2 b*h b=10cm h=10/2 sinTheta where theta is the angle between the two legs. area'=1/2 d (bh)/dtheta=1/2 b *cosTheta set area' to zero, then theta is 90 deg, so find largest area.
*Wednesday, September 2, 2009 at 10:09pm by bobpursley*

**Calculus**

1. Find the area of the region bounded by the curves and lines y=e^x sin e^x, x=0, y=0, and the curve's first positive intersection with the x-axis. 2. The area under the curve of y=1/x from x=a to x=5 is approximately 0.916 where 1<=a<5. Using your calculator, find a. 3...
*Wednesday, February 27, 2013 at 11:17am by Jessy152*

**Calculus**

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. x+y^2=42, x+y=0
*Thursday, November 21, 2013 at 10:29pm by Kelly*

**Math **

Determine the area of the region between the curves: f(x) = -3x^3 + 6.25x^2 + 27.1875x - 9.328125 and g(x) = 1x^2 + 3x -3
*Sunday, May 26, 2013 at 1:31pm by mark *

**calculus**

How do you find the ariea between these curves? y=4x^2 y=7x^2 4x+y=3 x>=0?
*Monday, October 24, 2011 at 11:42am by johnathon*

**math-calculus 2**

Consider the given curves to do the following. 64 y = x^3, y = 0, x = 4 Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about y = 1.
*Friday, June 21, 2013 at 5:23pm by Biff*

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