Use the transformation x = 3v + 1 and y = 2u + 2 and the Jacobin, to set up an integral, with...
Question:
Use the transformation {eq}\displaystyle x=3v+1 {/eq} and {eq}\displaystyle y = 2u+2 {/eq} and the Jacobin, to set up an integral, with limits of integration, to evaluate {eq}\displaystyle \int \int_{R} (x-1) (y-2)^2 e^{4(x-1)^2+9(y-2)^2} dA {/eq} where {eq}\displaystyle R {/eq} is the region bounded by the ellipse {eq}\displaystyle 4(x-1)^2 + 9(y-2)^2 = 36 {/eq}.
Integration:
Integration is used to find the value of the function {eq}f\left( x \right) {/eq} whose derivative {eq}D\left( {f\left( x \right)} \right) {/eq} is equal to the function {eq}f\left( x \right) {/eq}. The definite integral can be written in the form, {eq}\int\limits_a^b {f\left( x \right)} dx {/eq} where {eq}a {/eq} and {eq}b {/eq} are knows as the limits of integration.
Answer and Explanation: 1
Become a Study.com member to unlock this answer! Create your account
View this answerGiven:
- The transformation {eq}x = 3v + 1 {/eq} and {eq}y = 2u + 2 {/eq}.
Consider {eq}x = 3v + 1 {/eq} and {eq}y = 2u + 2 {/eq}.
Then, ...
See full answer below.
Ask a question
Our experts can answer your tough homework and study questions.
Ask a question Ask a questionSearch Answers
Learn more about this topic:
Get access to this video and our entire Q&A library
Integration Problems in Calculus | Examples & Solutions
from
Chapter 13 / Lesson 13
147K
Learn what integration problems are. Discover how to find integration sums and how to solve integral calculus problems using calculus example problems.
Related to this Question
- Use the transformation x=3v and y=2u and the Jacobian, to set up an integral, with limits of integration, to evaluate \iint_{R}^{2} x^{2} y^{2} d A where R is the region bounded by the ellipse 4
- Use the transformation x = \sqrt{2u} - \sqrt{2/3} v; y = \sqrt{2u} + \sqrt{2/3} v to evaluate the integral \iint_R (x^2-xy+y^2) \, dA where R is the region bounded by the ellipse x^2 -xy +
- Use the given transformation to evaluate the integral intint_Rx^2 dA, where R is the region bounded by the ellipse 9x^2+4y^2=36; x=2u, y=3y.
- Use the transformation x = \sqrt 5 u - \sqrt{5/7}v, \quad y = \sqrt 5 u + \sqrt{5/7}v to evaluate the integral \iint_R (2x^2 - 3xy + 2y^2 ) \,dA , where R is the region bounded by the ellipse
- Use the transformation x = 2u, y = 5v to evaluate the integral \iint_R 5x^2 \, dA where R is the region bounded by the ellipse 25x^2 + 4y^2 = 100
- Use the transformation x = 4u, y = 5v to evaluate the integral \iint_R 7x^2\, dA , where R is the region bounded by the ellipse 25x^2 + 16y^2 = 400
- Use the given transformation to evaluate the integral. int int_{R} 8 x^2 d A, where R is the region bounded by the ellipse 4 x^2 + 9 y^2 = 36 ; x = 3 u, y = 2 v
- L = \int \int_R (8x^2 - xy + 8y^2) dA; x = \sqrt {8/15} u - \sqrt {8/17} v, y = \sqrt {8/15} u + \sqrt {8/17} v Use the given transformation to evaluate the given integral, where R is the region bounded by the ellipse 8x^2 - xy + 8y^2 = 8.
- Use the given transformation to evaluate the integral. Int R(8x^2-5xy+8y^2)dA, where R is the region bounded by the ellipse 8x^2 - 5xy + 8y^2 = 2; x = \sqrt{2}/11u- \sqrt{2}/21v, y= \sqrt{2}/11u+ \s
- Use the given transformation to evaluate the integral. \int \int_R 4x^2 dA, where R is the region bounded by the ellipse 4x^2 + 9y^2 = 36; x = 3u, y = 2v.
- Use the given transformation to evaluate the integral. \int \int_R 6x^2 dA , where R is the region bounded by the ellipse 25x^2 + 9y^2 = 225; x = 3u, y = 5v.
- Use the given transformation to evaluate the integral. \int \int_R 5x^2 dA, where R is the region bounded by the ellipse 16x^2 + 25y^2 = 400; x = 5u, y = 4v
- Use the given transformation to evaluate the integral. \int \int_{R}8x^{2}dA, where R is the region bounded by the ellipse 16x^{2}+9y^{2}=144;x=3u,y=4v
- Use the given transformation to evaluate the integral. \iint_R (7x^2 - 3xy + 7y^2) dA ,where R is the region bounded by the ellipse 7x^2 -3xy + 7y^2 = 3; x = \sqrt{3/11} u -\sqrt{ 3/17} v, y = \sqrt{
- Use the given transformation to evaluate the integral int int_{R} (x^2 - x y + y^2) d A where R is the region bounded by the ellipse x^2 - x y + y^2 = 2 ; x = square root{2}u - square root{2/3}v ; y =
- Use the transformation x=3v+1 and y=2u+2 and the jacobian, to set up an integral, with limits of integration, to evaluate (x-1)(y-2)2e(4(x-1)^2)+9(y-2)^2dA where R is the region bounded by the ellip
- Use the given transformation to evaluate the given integral, where R is the region bounded by the ellipse 7x^2 - 3xy + 7y^2 = 10. L= integral integral_R (7x^2-3xy+7y^2)dA, x=(10/11)^{(1/2)}u-(10/17)^{
- Use the given transformation to evaluate the integral. \int\int_R4x^2dA where R is the region bounded by the ellipse 9x^2 + 16y^2 =144; x=4u, y = 3v
- Use the given transformation to evaluate the integral. \iint_{R} (2x^{2} - 3xy + 2y^{2}) dA, where R is the region bounded by the ellipse 2x^{2} - 3xy + 2y^{2}. x = \sqrt{3}u - \sqrt{3/7}v, y = \sqr
- Use the given transformation to evaluate the integral. \iint\limits_{R} 8x^2 \enspace dA where R is the region bounded by the ellipse 25x^2 + 36y^2 = 900; x = 6u, y = 5v
- Use the given transformation to evaluate the integral. \iint_R 8x^2 dA, where R is the region bounded by the ellipse 25x^2 + 9y^2 = 225; x = 3u, y = 5v
- Use the given transformation to evaluate the integral. Double integral over R of 6x^2 dA, where R is the region bounded by the ellipse 25x^2 + 4y^2 = 100; x = 2u, y = 5v.
- Use the given transformation to evaluate the integral. Double integral over R of 2x^2 dA, where R is the region bounded by the ellipse 9x^2 + 4y^2 = 36; x = 2u, y = 3v.
- Use the given transformation to evaluate the integral. Double integral over R of 6x^2 dA, where R is the region bounded by the ellipse 9x^2 + 4y^2 = 36; x = 2u, y = 3v.
- Use the given transformation to evaluate the integral. integral_R(8 x^{2} - 9 x y + 8 y^{2}) dA, where R is the region bounded by the ellipse 8 x^{2} - 9 x y + 8 y^{2} = 3; x = square root {3 / 7} u -
- Use the given transformation to evaluate the given integral, where R is the region bounded by the ellipse 25x^{2}+36y^{2}=900. L=\int \int_{R}10y^{2}dA; x=6u,y=5v
- Use the transformation x = 5u, y = 2v to evaluate the integral \iint_R 4x^2 \, dA, where R is the region bounded by the ellipse 4x^2 + 25y^2 = 100
- Use the transformation x = \sqrt{9/5}u - \sqrt{9/19}v, y = \sqrt{9/5}u + \sqrt{9/19}v to evaluate the integral \iint_R (6x^2 - 7xy + 6y^2) \,dA , where is the region bounded by the ellipse
- Use the given transformation to calculate the integral. Double integral over R of 2x^2 dA, where R is the region bounded by the ellipse 9x^2 + 4y^2 = 36; x = 2u, y = 3v.
- Use the given transformation to evaluate the given integral, where R is the region bounded by the ellipse 10x^2 - 3xy + 10y^2 = 10. L= \iint_R (10x^2- 3xy+ 10y^2)dA
- Use the given transformation to evaluate the given integral, where R is the region bounded by the ellipse 9x^2 + 4y^2 = 36. L = double integral over R of 3y^2 dA; x = 2u, y = 3v.
- Use the transformation x = 2u, y = 3v to evaluate the double integral over R of x^2 dA, where R is the region bounded by the ellipse 9x^2 + 4y^2 = 36.
- Use the transformation x = (sqrt(5) u - sqrt(5/3) v) and y = (sqrt(5) u + sqrt(5/3) v) to evaluate the double integral over R of (x^2 - xy + y^2) dA where R is the region bounded by the ellipse x^2 -
- Use the transformation x=3v+ 1 and y=2u+ 2 and the Jacobian, to set up an integral with limits of integration, to evaluate \iint_R (x-1)(y-2)^2 e^{4(x-1)^2+ 9(y-2)^2} dA where R is the region bounded
- Use the given transformation to evaluate the given integral, where R is the region bounded by the ellipse 4x^2 + 9y^2 = 36. \mathcal{L} = \iint_R 5 x^2 dA; x = 3u, y = 2v
- Use the given transformation to evaluate the given integral, where R is the region bounded by the ellipse 9x^2 - 5xy + 9y^2 = 3.
- Use the given transformation to evaluate the given integral, where R is the region bounded by the ellipse 9 x^2 - x y + 9 y^2 = 10. L = int int_{R} (9 x^2 - x y + 9 y^2) dA ; x = sqrt{10 / 17} u - sqr
- Use the given transformation to evaluate the given integral,where R is the region bounded by the ellipse 16x2 + 25y2 = 400. x = 5u, y =4v
- Use the given transformation to evaluate the given integral, where R is the region bounded by the ellipse 4x^2-5xy+4y^2=4. L=intint_R(4 x^2-5xy+4y^2) dA; x=sqrt4/3 u-sqrt4/13 v, y=sqrt4/3 u+sqrt4
- Use the given transformation to evaluate the integral. double integral_R (7 x^2 - 11 x y + 7 y^2) dA, where R is the region bounded by the ellipse 7 x^2 - 11 x y + 7 y^2 = 2; x = square root {2 / 3} u
- Use the given transformation to evaluate the integral. double integral_R 10 x^2 dA, where R is the region bounded by the ellipse 9 x^2 + 16 y^2 = 144; x = 4 u, y = 3 v.
- Use the given transformation to evaluate the integral. double integral_R {3 x^2} dA, where R is the region bounded by the ellipse 25 x^2 + 9 y^2 = 225; x=3 u, y=5 v.
- Use the given transformation to evaluate the integral. double integral_R 3 x^2 dA, where R is the region bounded by the ellipse 25 x^2 + 16 y^2 = 400; x = 4 u, y = 5 v
- Use the transformation x = 2u and y = 3v to evaluate \int\int_R x^2 \ dA where R is the region bounded by the ellipse 9x^2 + 4y^2 = 36.
- Use the given transformation to evaluate the given integral, where R is the region bounded by the ellipse \int \int (9x^{2} - 5xy + 9y^{2})dA . x = sqrt(5/13)u - sqrt(5/23)v, y = sqrt(5/13)u + sqrt(
- Use a Jacobian transformation to evaluate the double integral over R of x^2 dA, where R is the region bounded by the ellipse 16x^2+4y^2=64.
- Use a Jacobian transformation to evaluate double integral over R of x^2 dA, where R is the region bounded by the ellipse 16x^2 + 4y^2 = 64.
- Use the transformation x = u - sqrt(3)v; y = u + sqrt(3)v to evaluate the integral: double integral over E of y^2 dydx where E is the ellipse enclosed by x^2 + xy + y^2 = 3.
- Use the transformation x=3u, y = 5v to evaluate the integral \iint_R 10x^2 \,dA , where R is the region bounded by the ellipse 25x^2 + 9y^2 = 225
- Use an appropriate transformation to evaluate the integral I = ∫∫_Dsin(4x^2 + 16y^2)dxdy when D is the region bounded by the ellipse 4x^2 + 16y^2 = π.
- Use the transformation x = 3u, y = 5v to evaluate the integral \iint_R 18x^2 \, dA , where R is the region bounded by the ellipse 50x^2 + 18y^2 = 450
- Use the given transformation to evaluate the given integral, when R is the region bounded by the ellipse 5x^2 - 9xy + 5y^2 = 10. L = double integral_R (5x^2 - 9xy + 5y^2) dA; x = square root 10 u - sq
- Use the given transformation to evaluate the given integral, where R is the region bounded by the ellipse 5 x^2 - 3 x y + 5 y^2 = 9. L = double integral_R (5 x^2 - 3 x y + 5 y^2) dA; x = square root {
- Use the given transformation to evaluate the integral. \iint_{R} 9\sin(81x^{2} + 100y^{2}) dA, where R is the region in the first quadrant bounded by the ellipse 81x^{2} + 100y^{2} = 1; u = 9x, v = 10y
- Use a Jacobian transformation to calculate the double integral over R of x^2 dA, where R is the region bounded by the ellipse 16x^2 + 4y^2 = 64.
- Let R be the region bounded by the graph of the ellipse x^2 - xy + y^2 = 2. Using the change of variables transformation x = \sqrt 2u - \sqrt {\frac {2}{3v and y = \sqrt 2u + \sqrt {\frac {2}{3v,
- Use the given transformation evaluate the given interval, where R is region bounded by the ellipse 5x^2 - 9xy + 5y^2 = 3. L = double integral_R (5x^2 - 9xy + 5y^2) dA; x = square root 3 u - square roo
- Use the given transformation to evaluate the given integral, where R is the region bounded by the ellipse 25x^2 + 4y^2 = 100. L = double integral_R 3x^2 dA; x = 2u, y = 5v.
- Use the transformation x = 3v + 1 and y = 2u + 2 and the Jacobian, to set up an integral, with limits of integration, to evaluate Int Int(x-1)(y-2)^2 e^{4(x-1)^2 + 9(y-2)^2 dA where R is the region bo
- Use the given transformation to evaluate the integral \iint_R 6x^2 dA where R is the region bounded by the ellipse 9x^2 + 36y^2 = 324, x = 6u, y = 3v.
- Use the given transformation to evaluate the given integral, where R is the region bounded by the ellipse 4x^2 - 3xy + 4y^2 = 4. L = double integral_R (4x^2 - 3xy + 4y^2) dA; x = square root of {4 / 5} u - square root of {4 / 11}v, y = square root of {4 /
- Evaluate the integral \iint_R 2\sin(16x^2+49y^2) \,dA by making an appropriate change of variables, where R is the region in the first quadrant bounded by the ellipse 16x^2 + 49y^2 =1
- Evaluate the integral \iint_R 8 \sin(25x^2 + 64y^2) \,dA by making an appropriate change of variables, where R is the region in the first quadrant bounded by the ellipse 25x^2 + 64y^2 = 1
- Use the given integration to evaluate the given integral, where R is the region bounded by the ellipse 4x^2 + 3xy + 4y^2 = 2. L = double integral_R (4x^2 + 3xy + 4y^2) dA; x = square root {2 / 5} u -
- Use the given transformation to evaluate the integral. \iint_R 7x^2 dA, where R is the region bounded by the ellipse 25x^2 + 36y^2 = 900; x = 6u, y = 5v.
- Evaluate the integral by making an appropriate change of variables. \iint_R 6\sin(25x^2 + 81y^2) dA where R is the region in the first quadrant bounded by the ellipse 25x^2 + 81y^2 = 1.
- Evaluate the integral by making an appropriate change of variables. Integral integral_R 7 sin(9x^2 + 36y^2) dA, where R is the region in the first quadrant bounded by the ellipse 9x^2 + 36y^2 = 1
- Use the transformation x = 3u, y = 5v to evaluate the integral \iint_R 9x^2 \, dA , where R is the region bounded by the ellipse 25x^2 + 9y^2 = 225 .
- Let R be the region bounded by the ellipse 4x^2 + 25y^2 = 400 and T be the transformation T(u,v) = (x,y) given by x=5u, y = 2v . Find the Jacobian of T.
- Evaluate the following integrals double integral_R (x y+2) dA, where R is the region bounded by the ellipse x^2 + 9 y^2 = 18.
- A) Evaluate the integral by making an appropriate change of variables. Double integral over R of 8sin(25x^2 + 9y^2) dA, where R is the region in the first quadrant bounded by the ellipse 25x^2 + 9y^2
- Use the given transformation to evaluate the given integral, where R is the region bounded by the elipse 7x^2 - xy+7y^2=1. \hspace{10mm} L = \int \int_{R} (7x^2 -xy+7y^2) dA; \hspace{5mm} x= \
- Set-up an evaluate an integral for the area of the ellipse \frac{(x + 1)^{2{3} + \frac{(y - 3)^{2{6} = 1. Evaluate the integral or show it diverges: \int_{0}^{\infty} \frac{x}{e^x} dx
- Evaluate the integral by making an appropriate change of variable. double integral cos(4x^2+9x^2) dA, where R is the region in the first quadrant bounded by the ellipse
- Evaluate the integral by making an appropriate change of variables. \iint_{R} 8\sin(25x^{2} + 4y^{2}) dA, where R is the region in the first quadrant bounded by the ellipse 25x^{2} + 4y^{2} = 1
- Evaluate the integral by making an appropriate change of variables. int int_R 6 sin(81x^2 + 81y^2) dA, where R is the region in the first quadrant bounded by the ellipse 81x^2 + 81y^2 = 1
- Evaluate the integral by making an appropriate change of variables. Double integral over R of 5sin(49x^2 + 16y^2) dA, where R is the region in the first quadrant bounded by the ellipse 49x^2 + 16y^2 =
- Evaluate the integral by making an appropriate change of variables. Double integral over R of 2sin(9x^2 + 49y^2) dA, where R is the region in the first quadrant bounded by the ellipse 9x^2 + 49y^2 = 1
- Evaluate the integral by making an appropriate change of variables. Double integral over R of 3sin(9x^2 + 25y^2) dA, where R is the region in the first quadrant bounded by the ellipse 9x^2 + 25y^2 = 1
- Evaluate the integral by making an appropriate change of variables. Double integral over R of 8sin(49x^2 + 4y^2) dA, where R is the region in the first quadrant bounded by the ellipse 49x^2 + 4y^2 = 1
- Use the given transformation to evaluate the integral. int int_R (7x^2-5xy + 7y^2) dA , where R is the region bounded by the ellipse 7x^2 - 5xy + 7y^2 =2; x=square root2/9u - square root 2/19v, y
- Use the transformation u = xy, v = xy^2 to evaluate the integral \iint_R 10y^2 \,dA , where R is the region bounded by the curves xy=6,xy=9 ,xy^2 = 6, xy^2 =9
- Evaluate the given integral by making an appropriate change of variables, where R is the region in the first quadrant bounded by the ellipse 9x^2 + 100y^2 = 1. L = Double integral 5sin(36x^2 + 400y^2) dA
- Evaluate the given integral by making an appropriate change of variables, where R is the region in the first quadrant bounded by the ellipse 100x^2 + 49y^2 = 1. Double integral 2sin(400x^2 + 196y^2) dA
- Use change of variable: (x = 2u, y = 3v) to evaluate the double integral over R of x^2 dA, where R is the region bounded by the ellipse 9x^2 + 4y^2 = 36.
- Use the given transformation to evaluate the integral. Double integral over R of 9xy dA, where R is the region in the first quadrant bounded by the lines y = 2/3 x and y = 3/2 x and the hyperbolas xy
- (a) Use the given transformation to evaluate the integral double integral 4xy dA where R is the region in the first quadrant bounded by the lines y = 1/2x, y = 2x and the hyperbolas xy = 1/2 and xy =
- Use the given transformation to evaluate the integral. double integral R 5xy dA, where R is the region in the first quadrant bounded by the lines y = (1/3)x and y = (3/2)x and the hyperbolas xy = 1/3
- Use the transformation G(u,v) = (\frac{u}{v+1}, \frac{uv}{v+1}) to evaluate the integral \iint_D (x+y) \,dx\,dy , where D is the region bounded by y=3-x, y=6-x, y=2x,y=x
- Transform the integral \int_R \int xy^3 dA where R is the region bounded by the curves y = x, y = 4x, xy = 2, xy = 5 and the transformation is T: x = \frac{u}{v}, y = v.
- Evaluate the integral by making an appropriate change of variables. \int \int_R 6 \sin(49x^2 + 36y^2)dA, where R is the region in the first quadrant bounded by the ellipse 49x^2 + 36y^2 = 1.
- Use the transformation: x = \frac{u}{v}, y = v, to evaluate the integral \iint_Rxy dA, where: R is the region in the first quadrant, bounded by the lines y = x, y = 3x and the hyperbolas xy = 1, xy = 3.
- Use the given transformation to evaluate the integral. \iiint_R 10x^2 dA where R is the region bounded by the ellipse 4x^2 + 9y^2 = 36; x = 3u, y = 2v
- Use the given transformation to evaluate the double integral: intint_D(x^2+y^2) dA where D is the ellipse with equation: 3x^2+17y^2=51. Let x=sqrt17 u and y=sqrt3 v, now evaluate the same integral
- Use a line integral to compute the area of the region bounded by the ellipse 16x^2 + 4y^2 = 64.
- Use the transformation u = xy, v=xy^2 to evaluate the integral \iint_R 4y^2 \,dA , where R is the region bounded by the curves xy=5,xy=6,xy^2=5,xy^2=6
- Let E be the interior of the ellipse defined by (1/9)(x-3)2+(1/16)(y+2)2=1. Use the change of variables x = 3vcos(u)+3 and y=4vsin(u) to evaluate double integrate over E (1/sqrt(16x2-96x+9y2+36y+180))dxdy.
- Use the given transformation to evaluate the integral. Double integral over R of 10x^2 dA, where R is the region bounded by the ellipse 16x^2 + 9y^2 = 144; x = 3u, y = 4v.
- Use the transformation, x = u / v , y = u v to evaluate the double integral D x y d A where D is the region bounded by the hyperbolas, x y = 1 , x y = 5 and the lines y = x , y = 3 x in the first quadrant.
- Use the given transformation to evaluate the integral, \iint_{R} 9xy dA, where R is the region in the first quadrant bounded by the lines y = \frac{2}{3}x and y = \frac{3}{2}x and the hyperbolas xy = \frac{2}{3} and xy = \frac{3}{2}; x = \frac{u}{v}
Explore our homework questions and answers library
Browse
by subject