Thursday

April 24, 2014

April 24, 2014

Number of results: 43,450

**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) = 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**

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 (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 **

Sketch the region enclosed by the given curves. y = 4/X y = 16x, y = 1X/16 x > 0 and the area between the curves
*Saturday, April 12, 2014 at 3:13pm by LAUERN PLEASE HELP *

**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*

**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 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)**

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*

**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**

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**

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 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 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**

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 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**

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 - 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*

**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 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*

**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 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 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*

**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 - 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*

**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**

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. Between y = −x and y = x/2 for x in [0, 6]
*Thursday, April 12, 2012 at 10:32pm by Alex*

**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 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**

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*

**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 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**

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**

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**

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*

**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*

**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*

**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 **

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 *

**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 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 **

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?*

**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 *

**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**

3) Find the area bounded by the curves f(x)= x^3 + x^2 and g(x)= 2x^2 + 2x. I get 4653pi/105 after help from Mr. Reiny, but this is wrong. Have I calculated incorrectly? Thank you
*Wednesday, May 22, 2013 at 12:29am by Hannah*

**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 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 = 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*

**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*

**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 - 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 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**

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 PLEASE HELP **

The birth rate of a population is b(t) = 2300e0.024t people per year and the death rate is d(t)= 1450e0.018t people per year, find the area between these curves for 0 ≤ t ≤ 10
*Saturday, April 12, 2014 at 4:05pm by LAUERN PLEASE HELP *

**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 which is bounded by the polar curves theta =pi and r=2theta 0<theta<1.5pi inclusive
*Thursday, February 4, 2010 at 5:31pm by Salman*

**Calculus**

Find the area of the region which is bounded by the polar curves theta =pi and r=2theta 0<theta<1.5pi inclusive
*Friday, February 5, 2010 at 7:19am by Salman*

**Calculus**

Find the area of the region which is bounded by the polar curves theta =pi and r=2theta 0<theta<1.5pi inclusive
*Friday, February 5, 2010 at 3:18pm by Salman*

**Calculus**

Find the area of the region which is bounded by the polar curves theta =pi and r=2theta 0<theta<1.5pi inclusive
*Saturday, February 6, 2010 at 4:17am by Salman*

**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**

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=3(x^(1/2)) , y=5 and 2y+3x=6
*Friday, October 26, 2012 at 9:16pm by Kathy*

**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*

**Calculus-check my answers please**

dy/dx= (y^2 -1)/x 1. Give the general equation of the curves that satisfy this equation. 2. Show that the straight lines y=1 and y=-1 are also solutions 3. Do any of the curves you found in 1) intersect y=1? My Ans: 1. The general solution i found out to be x^2 + C = (y-1)/(y+...
*Tuesday, November 7, 2006 at 3:45pm by Hebe*

**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*

**calculus**

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area S of the region. y=sqrt(x) , y=1/2 x , x=25
*Friday, January 30, 2009 at 6:53pm by ken*

**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=3sqrt(x) , y=3 and 2y+3x=6.
*Friday, October 22, 2010 at 3:55pm by michael*

**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=3sqrt(x) , y=3 and 2y+3x=6.
*Friday, October 22, 2010 at 3:57pm by michael*

**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=3sqrt(x) , y=3 and 2y+3x=6.
*Friday, October 22, 2010 at 4:02pm by michael*

**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=3sqrt(x) , y=3 and 2y+3x=6.
*Friday, October 22, 2010 at 4:13pm by michael*

**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=3sqrt(x) , y=3 and 2y+3x=6.
*Friday, October 22, 2010 at 4:14pm by michael*

**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=3sqrt(x) , y=3 and 2y+3x=6.
*Friday, October 22, 2010 at 4:54pm by michael*

**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=3sqrt(x) , y=3 and 2y+3x=6.
*Friday, October 22, 2010 at 5:15pm by michael*

**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=3sqrt(x) , y=3 and 2y+3x=6.
*Friday, October 22, 2010 at 5:24pm by michael*

**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=5 rootx, y=5, and 2y+2x=7.
*Sunday, April 24, 2011 at 9:18pm by Amy*

**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=3sqrtx and y=3 and 2y+2x=5
*Tuesday, August 16, 2011 at 8:18pm by Brit*

**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=3sqrtx , y=3 , 2y+2x=5
*Wednesday, August 17, 2011 at 1:49am by Dan*

**Calculus **

1. Find the are between the curves y=e^x and y=4-x^2 graphically. a.) set up the integral b.) include bounds rounded to three decimal places c.) use integral function on calculator 2. Find the area of the "triangular" region bounded by y=4-x on the left, y=sqrt(x-2) on the ...
*Thursday, April 14, 2011 at 8:51pm by Amy*

**CALCULUS**

Sketch the region enclosed by the given curves. y = tan 3x, y = 2 sin 3x, −π/9 ≤ x ≤ π/9 then then find the area. i can sketch but cant find correct area
*Saturday, April 12, 2014 at 4:06pm by LAUERN PLEASE HELP *

**Calculus: Area **

Can someone look at my work and see what i did wrong. I did this 100 times but i keep getting it wrong 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=3x^(1/2) , y=4 and 2y+1x=4 y=3(x^(1/2...
*Saturday, April 28, 2012 at 6:48pm by jasmineT*

**calculus**

Sketch the region in the first quadrant enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. y=10cosx, y=10sin2x,x=0
*Thursday, April 8, 2010 at 4:19pm by lssa*

**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 WXMAXIMA !!PLEASE! **

y=log(100493/40000*x-1) y=atan(100493/40000*x+1) y=exp(-100493/40000*x+1) Note you will need 2 integrals to represent the area. Approximate the area of the region bounded by the given curves. I can do it with two but kinda struggling over here please helppp!
*Wednesday, March 7, 2012 at 4:39pm by Jordan *

**calculus**

Find the area between the graph of f and the x-axis over the interval [−4, 1] when f(x) = 4 + 4x. 1. Area = 34 sq. units 2. Area = 30 sq. units 3. Area = 28 sq. units 4. Area = 32 sq. units 5. Area = 26 sq. units
*Sunday, February 26, 2012 at 2:58pm by Hillcrest*

**Calculus**

Find the centroid of the area bounded by the curves: sqrt(x)+sqrt(y)=sqrt(a) x=0 y=0
*Monday, November 14, 2011 at 10:15pm by Shayna*

**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=5x^(1/2) , y=4 and 2y+1x=6 I've been trying this problem for about 3 hours. Please help!!!!!
*Tuesday, July 10, 2012 at 4:05pm by Sammy*

**calculus**

Find the number b such that the line y = b divides the region bounded by the curves y = 16x2 and y = 9 into two regions with equal area. (Round your answer to two decimal places.)
*Wednesday, November 9, 2011 at 1:42pm by Courtney*

**calculus**

Find the number b such that the line y = b divides the region bounded by the curves y = 16x2 and y = 9 into two regions with equal area. (Round your answer to two decimal places.)
*Wednesday, November 9, 2011 at 1:43pm by Courtney*

**calculus**

The figure below shows the curves y=square root of x, x=9, y=0 and a rectangle with the sides parallel to the axes and its left end at x=a. Find the dimensions of the rectangle having the maximum possible area.
*Sunday, November 13, 2011 at 2:28pm by Anonymous*

**Calculus**

Find the number b such that the line y = b divides the region bounded by the curves y = 4x2 and y = 1 into two regions with equal area. (Round your answer to two decimal places.)
*Monday, November 18, 2013 at 8:45pm by Graham*

**calculus**

Sketch the region bounded by the curves y = x^2, y = x^4. 1) Find the area of the region enclosed by the two curves; 2) Find the volume of the solid obtained by rotating the above region about the x-axis; 3) Find the volume of the solid obtained by rotating the above region ...
*Thursday, June 9, 2011 at 8:46pm by Megan*

**Calculus**

Can someone look over my work and tell me if my steps look correct? I'm trying to correct some problems that looked wrong. Instructions: Find the total areas between the given curves. 1. x= (y^3) and x=(y^2) on the interval [0,1] (integral from 0 to 1 of) ((y^3)-(y^2))dy = (...
*Thursday, September 10, 2009 at 7:47pm by Jenna*

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