Search: Calculus (Area Between Curves)

Number of results: 67,417

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)
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

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

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.
Thursday, January 24, 2013 at 7:46pm 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 Curves)
Thank you, I was just really unsure of my answer!
Wednesday, February 29, 2012 at 4:21pm by Mishaka

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/...
Wednesday, February 29, 2012 at 2:39pm by Steve

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 f(x)= ln(x+1)+2 g(x)= x^2-8x+12
Tuesday, December 23, 2008 at 5:38pm by mary

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 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

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. ...
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 ...
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 (...
Tuesday, December 11, 2012 at 12:54pm 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 ...
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
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 ...
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 ...
Thursday, November 25, 2010 at 10:12pm by K

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
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 - 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 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 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 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 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

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 "...
Thursday, January 29, 2009 at 5:22am by drwls

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

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.

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 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 - 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

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: 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

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
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
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 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 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
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

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
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
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
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 ...
Wednesday, February 27, 2013 at 11:17am by Jessy152

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

Physics - Mechanics and Ocsillatory Motion
find the area of the figure from x1 to x2. That area is the energy in joules added. Then set that area to difference in KE (KEf-KEi) use basic area under the curves (break it up into rectangles and add).
Saturday, October 30, 2010 at 6:42pm by bobpursley

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-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

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) ...
Thursday, April 14, 2011 at 8:51pm by Amy

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

Calculus: Area
go to wolframalpha . com and enter graph x=4y^2/9,x+2y=4,y=4,y=3/2,x=1 to see the curves involved. The area of interest is a little v-shaped chunk from (-4,4) down to (1,3/2) and back up to (64/9,4) If you integrate on x, then the integrands use the limits you have, but the ...
Saturday, April 28, 2012 at 6:48pm by Steve

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 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(Lab)
Well, first graph the graph of f(x)=-1/10x^2 + 3 2. We are going to approximate the area between f and the x-axis from x = 0 to x = 4 using rectangles (the method of Riemann sums). This is not the entire area in the first quadrant, just most of it. Draw four inscribed ...
Wednesday, January 25, 2012 at 5:32pm by Yoona

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

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

maths
y = x^2 = sqrt(8x) x^4 = 8x (x^3 - 8)*x = 0 x = 0 or 2 For the area between the curves, integrate (8x)^(1/2) - x^2 from 0 to 2. sqrt8*2^(3/2)/(3/2) - (2^3)/3 = 16/3 - 8/3 = 8/3
Monday, February 7, 2011 at 1:19am by drwls

Calculus
Please help? Let F(a) be the area between the x-axis and the graph of y=x^2cos(x/4) between x=0 and x=a, for a>0 (consider the area to be negative if the graph lies below the axis). Find the formula for F(a).
Thursday, September 15, 2011 at 9:00pm by Catherine

Calculus
Please help? Let F(a) be the area between the x-axis and the graph of y=x^2cos(x/4) between x=0 and x=a, for a>0 (consider the area to be negative if the graph lies below the axis). Find the formula for F(a).
Wednesday, September 14, 2011 at 9:16pm by Claudia

Calculus
Let F(a) be the area between the x-axis and the graph of y=x^2cos(x/4) between x=0 and x=a, for a>0 (consider the area to be negative if the graph lies below the axis). Find the formula for F(a).
Wednesday, September 14, 2011 at 6:28pm by Catherine

Calculus
Let F(a) be the area between the x-axis and the graph of y=x^2cos(x/4) between x=0 and x=a, for a>0 (consider the area to be negative if the graph lies below the axis). Find a formula for F(a).
Friday, September 9, 2011 at 9:41pm by Catherine

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
Let F(a) be the area between the x-axis and the graph of y=x^2cos(x/4) between x=0 and x=a, for a>0 (consider the area to be negative if the graph lies below the axis).
Saturday, September 10, 2011 at 1:24am by Catherine

Calculus (Area Between Curves)
Just answered a minute ago at: http://www.jiskha.com/display.cgi?id=1330550488
Wednesday, February 29, 2012 at 4:43pm by MathMate

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 = 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
a.) Given y=square root of (x) and y= (x) Find the area between two curves from x=0 to x=1 by integrating with respect to x (top-bottom) I'm more concerned about this one though: b.) Now integrate with respect to y given between y=0 and y=1 Do I just graph it on my ...
Thursday, May 12, 2011 at 9:56am by Jill

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 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
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
Thursday, February 4, 2010 at 5:31pm by Salman

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