Question 1

Use the Midpoint Rule with four squares of equal size to estimate the double integral.  ∬Rcos⁡(x4+y4) dA,R={(x,y) ∣0≤x≤0.5,0≤y≤0.5}\iint _ { R } \cos \left( x ^ { 4 } + y ^ { 4 } \right)  d A , R = \{ ( x , y )  \mid 0 \leq x \leq 0.5,0 \leq y \leq 0.5 \}∬R​cos(x4+y4) dA,R={(x,y) ∣0≤x≤0.5,0≤y≤0.5}

Accepted Answer

A)    0.5750.5750.575 
B)    0.66220.66220.6622 
C)    0.24990.24990.2499 
D)    0.47210.47210.4721 
E)    0.76220.76220.7622F) A) and C)
G) B) and E)

Question 2

Calculate the iterated integral. $$\int _ { 1 } ^ { 4 } \int _ { 0 } ^ { 3 } ( 3 + 4 x y ) d x d y$$

Accepted Answer

The answer of Calculate the iterated integral. \[\int _ {...

Question 3

Sketch the solid bounded by the graphs of the equations $z = x ^ { 2 } + y ^ { 2 }$ and $z = 50 - x ^ { 2 } - y ^ { 2 }$, and then use a triple integral to find the volume of the solid.

Accepted Answer

The answer of Sketch the solid bounded by the graphs...

Question 4

$\text { Evaluate } \iint _ { D } 4 x ^ { 2 } y ^ { 2 } d A \text { where } D \text { is the figure bounded by } y = 1 , y = 2 , x = 0 \text { and } x = y \text {. }$

Accepted Answer

The answer of \(\text { Evaluate } \iint _ {...

Question 5

Calculate the iterated integral.Round your answer to two decimal places.  ∫06∫034x+4ydxdy\int _ { 0 } ^ { 6 } \int _ { 0 } ^ { 3 } \sqrt { 4 x + 4 y } d x d y∫06​∫03​4x+4y​dxdy

Accepted Answer

A)    94.2694.2694.26 
B)    134.26134.26134.26 
C)    114.26114.26114.26 
D)    84.2684.2684.26 
E)    74.2674.2674.26F) C) and D)
G) A) and B)

Question 6

Find the area of the surface.
The part of the surface  z=25−x2−y2z = 25 - x ^ { 2 } - y ^ { 2 }z=25−x2−y2  that lies above the  xyx yxy -plane.

Accepted Answer

A)    (626−626) ( \sqrt { 626 } - 626 ) (626c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​−626)  
B)    275(626626+1) \frac { 2 } { 75 } ( 626 \sqrt { 626 } + 1 ) 752​(626626c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​+1)  
C)    275π(626626−1) \frac { 2 } { 75 } \pi ( 626 \sqrt { 626 } - 1 ) 752​π(626626c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​−1)  
D)    275(626+1) \frac { 2 } { 75 } ( \sqrt { 626 } + 1 ) 752​(626c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​+1)  
E)    π(626626−1) \pi ( 626 \sqrt { 626 } - 1 ) π(626626c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​−1) F) A) and D)
G) C) and D)

Question 7

Find the center of mass of the lamina of the region shown if the density of the circular lamina is four times that of the rectangular lamina.

Accepted Answer

The answer of Find the center of mass of the...

Question 8

Use cylindrical coordinates to evaluate  ∭Tx2+y2dV\iiint _ { T } \sqrt { x ^ { 2 } + y ^ { 2 } } d V∭T​x2+y2​dV , where  TTT  is the solid bounded by the cylinder  x2+y2=1x ^ { 2 } + y ^ { 2 } = 1x2+y2=1  and the planes  z=4z = 4z=4  and  z=8z = 8z=8 .

Accepted Answer

A)    83π\frac { 8 } { 3 } \pi38​π 
B)    4π4 \pi4π 
C)    32π32 \pi32π 
D)    48π48 \pi48πE) A) and B)
F) All of the above

Question 9

Calculate the iterated integral. $$\int _ { 0 } ^ { 1 } \int _ { 0 } ^ { y } 5 \cos \left( y ^ { 2 } \right) d x d y$$

Accepted Answer

The answer of Calculate the iterated integral. \[\int _ {...

Question 10

Evaluate the integral by changing to polar coordinates. $\iint _ { D } e ^ { - x ^ { 2 } - y ^ { 2 } } d A$
$D$ is the region bounded by the semicircle $x = \sqrt { 9 - y ^ { 2 } }$ and the $y$-axis.

Accepted Answer

The answer of Evaluate the integral by changing to polar...

Question 11

Find the Jacobian of the transformation.  x=5αsin⁡β,y=4αcos⁡βx = 5 \alpha \sin \beta , y = 4 \alpha \cos \betax=5αsinβ,y=4αcosβ

Accepted Answer

A)    ∂(x,y) ∂(α,β) =9α\frac { \partial ( x , y )  } { \partial ( \alpha , \beta )  } = 9 \alpha∂(α,β) ∂(x,y) ​=9α 
B)    ∂(x,y) ∂(α,β) =36α\frac { \partial ( x , y )  } { \partial ( \alpha , \beta )  } = 36 \alpha∂(α,β) ∂(x,y) ​=36α 
C)    ∂(x,y) ∂(α,β) =−20α\frac { \partial ( x , y )  } { \partial ( \alpha , \beta )  } = - 20 \alpha∂(α,β) ∂(x,y) ​=−20α 
D)    ∂(x,y) ∂(α,β) =−20αsin⁡βcos⁡β\frac { \partial ( x , y )  } { \partial ( \alpha , \beta )  } = - 20 \alpha \sin \beta \cos \beta∂(α,β) ∂(x,y) ​=−20αsinβcosβ 
E)    ∂(x,y) ∂(α,β) =−α\frac { \partial ( x , y )  } { \partial ( \alpha , \beta )  } = - \alpha∂(α,β) ∂(x,y) ​=−αF) B) and E)
G) None of the above

Question 12

Use polar coordinates to find the volume of the solid under the paraboloid  z=x2+y2z = x ^ { 2 } + y ^ { 2 }z=x2+y2  and above the disk  x2+y2≤9x ^ { 2 } + y ^ { 2 } \leq 9x2+y2≤9 .

Accepted Answer

A)    81π81 \pi81π 
B)    162π162 \pi162π 
C)    324π324 \pi324π 
D)    27π27 \pi27π 
E)    40.5π40.5 \pi40.5πF) C) and D)
G) A) and C)

Question 13

Use spherical coordinates.
Evaluate  ∭B(x2+y2+z2) 2dV\iiint _ { B } \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right)  ^ { 2 } d V∭B​(x2+y2+z2) 2dV , where  BBB  is the ball with center the origin and radius 4 .

Accepted Answer

A)    655367π\frac { 65536 } { 7 } \pi765536​π 
B)    43747\frac { 4374 } { 7 }74374​ 
C)    43747π\frac { 4374 } { 7 } \pi74374​π 
D)    5598727π\frac { 559872 } { 7 } \pi7559872​π 
E)   None of these F) A) and B)
G) B) and E)

Question 14

Evaluate the double integral by first identifying it as the volume of a solid.  ∬R(15−2x) dA,R={(x,y) ∣2≤x≤5,3≤y≤8}\iint _ { R } ( 15 - 2 x )  d A , R = \{ ( x , y )  \mid 2 \leq x \leq 5,3 \leq y \leq 8 \}∬R​(15−2x) dA,R={(x,y) ∣2≤x≤5,3≤y≤8}

Accepted Answer

A)   300
B)   -100
C)   0
D)   200
E)   100 F) A) and E)
G) D) and E)

Question 15

Use the given transformation to evaluate the integral.
 ∬RxydA\iint _ { R } x y d A∬R​xydA , where  RRR  is the region in the first quadrant bounded by the lines  y=x,y=3xy = x , y = 3 xy=x,y=3x  and the hyperbolas  xy=2,xy=4;x=uv,y=vx y = 2 , x y = 4 ; x = \frac { u } { v } , y = vxy=2,xy=4;x=vu​,y=v .

Accepted Answer

A)    9.4479.4479.447 
B)    3.2963.2963.296 
C)    8.8418.8418.841 
D)    4.4474.4474.447 
E)    5.0885.0885.088F) C) and E)
G) A) and E)

Question 16

Calculate the double integral.  ∬R(9x2y3−15x4) dA,R={(x,y) ∣0≤x≤1,0≤y≤4}\iint _ { R } \left( 9 x ^ { 2 } y ^ { 3 } - 15 x ^ { 4 } \right)  d A , R = \{ ( x , y )  \mid 0 \leq x \leq 1,0 \leq y \leq 4 \}∬R​(9x2y3−15x4) dA,R={(x,y) ∣0≤x≤1,0≤y≤4}

Accepted Answer

A)   250
B)   230
C)   190
D)   180
E)   210 F) B) and D)
G) A) and E)

Question 17

Find the center of mass of a lamina in the shape of an isosceles right triangle with equal sides of length  a=25a = 25a=25  if the density at any point is proportional to the square of the distance from the vertex opposite the hypotenuse. Assume the vertex opposite the hypotenuse is located at  (0,0) ( 0,0 ) (0,0)  , and that the sides are along the positive axes.

Accepted Answer

A)    (10,10) ( 10,10 ) (10,10)  
B)    (10,25) ( 10,25 ) (10,25)  
C)    (25,25) ( 25,25 ) (25,25)  
D)    (5,10) ( 5,10 ) (5,10)  
E)   None of these F) D) and E)
G) A) and C)

Question 18

Use spherical coordinate to find the volume above the cone $z ^ { 2 } = x ^ { 2 } + y ^ { 2 }$ and inside sphere $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 2 a z$.

Accepted Answer

The answer of Use spherical coordinate to find the volume...

Question 19

Calculate the double integral.  ∬R(6x2y3−10x4) dA,R={(x,y) ∣0≤x≤1,0≤y≤4}\iint _ { R } \left( 6 x ^ { 2 } y ^ { 3 } - 10 x ^ { 4 } \right)  d A , R = \{ ( x , y )  \mid 0 \leq x \leq 1,0 \leq y \leq 4 \}∬R​(6x2y3−10x4) dA,R={(x,y) ∣0≤x≤1,0≤y≤4}

Accepted Answer

A)   190
B)   170
C)   130
D)   150
E)   120 F) A) and B)
G) A) and E)

Question 20

Evaluate the double integral by first identifying it as the volume of a solid.  ∬R(15−2x) dA,R={(x,y) ∣3≤x≤7,3≤y≤5}\iint _ { R } ( 15 - 2 x )  d A , R = \{ ( x , y )  \mid 3 \leq x \leq 7,3 \leq y \leq 5 \}∬R​(15−2x) dA,R={(x,y) ∣3≤x≤7,3≤y≤5}

Accepted Answer

A)   40
B)    −160- 160−160 
C)    −60- 60−60 
D)   140
E)   240 F) A) and E)
G) B) and E)

Exam 15: Multiple Integrals

Use spherical coordinates. Evaluate $\iiint _ { B } \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { 2 } d V$ , where $B$ is the ball with center the origin and radius 4 .

Correct Answer
verified

Find the center of mass of the lamina of the region shown if the density of the circular lamina is four times that of the rectangular lamina.

Correct Answer
verified

Calculate the double integral. $\iint _ { R } \left( 9 x ^ { 2 } y ^ { 3 } - 15 x ^ { 4 } \right) d A , R = \{ ( x , y ) \mid 0 \leq x \leq 1,0 \leq y \leq 4 \}$

Correct Answer
verified

Evaluate the double integral by first identifying it as the volume of a solid. $\iint _ { R } ( 15 - 2 x ) d A , R = \{ ( x , y ) \mid 2 \leq x \leq 5,3 \leq y \leq 8 \}$

Correct Answer
verified

Use the given transformation to evaluate the integral. $\iint _ { R } x y d A$ , where $R$ is the region in the first quadrant bounded by the lines $y = x , y = 3 x$ and the hyperbolas $x y = 2 , x y = 4 ; x = \frac { u } { v } , y = v$ .

Correct Answer
verified

$\text { Evaluate } \iint _ { D } 4 x ^ { 2 } y ^ { 2 } d A \text { where } D \text { is the figure bounded by } y = 1 , y = 2 , x = 0 \text { and } x = y \text {. }$

Correct Answer
verified

Use polar coordinates to find the volume of the solid under the paraboloid $z = x ^ { 2 } + y ^ { 2 }$ and above the disk $x ^ { 2 } + y ^ { 2 } \leq 9$ .

Correct Answer
verified

Evaluate the integral by changing to polar coordinates. $\iint _ { D } e ^ { - x ^ { 2 } - y ^ { 2 } } d A$ $D$ is the region bounded by the semicircle $x = \sqrt { 9 - y ^ { 2 } }$ and the $y$ -axis.

Correct Answer
verified

Calculate the double integral. $\iint _ { R } \left( 6 x ^ { 2 } y ^ { 3 } - 10 x ^ { 4 } \right) d A , R = \{ ( x , y ) \mid 0 \leq x \leq 1,0 \leq y \leq 4 \}$

Correct Answer
verified

Calculate the iterated integral.Round your answer to two decimal places. $\int _ { 0 } ^ { 6 } \int _ { 0 } ^ { 3 } \sqrt { 4 x + 4 y } d x d y$

Correct Answer
verified

Find the Jacobian of the transformation. $x = 5 \alpha \sin \beta , y = 4 \alpha \cos \beta$

Correct Answer
verified

Use cylindrical coordinates to evaluate $\iiint _ { T } \sqrt { x ^ { 2 } + y ^ { 2 } } d V$ , where $T$ is the solid bounded by the cylinder $x ^ { 2 } + y ^ { 2 } = 1$ and the planes $z = 4$ and $z = 8$ .

Correct Answer
verified

Find the area of the surface. The part of the surface $z = 25 - x ^ { 2 } - y ^ { 2 }$ that lies above the $x y$ -plane.

Correct Answer
verified

Evaluate the double integral by first identifying it as the volume of a solid. $\iint _ { R } ( 15 - 2 x ) d A , R = \{ ( x , y ) \mid 3 \leq x \leq 7,3 \leq y \leq 5 \}$

Correct Answer
verified

Sketch the solid bounded by the graphs of the equations $z = x ^ { 2 } + y ^ { 2 }$ and $z = 50 - x ^ { 2 } - y ^ { 2 }$ , and then use a triple integral to find the volume of the solid.

Correct Answer
verified

Correct Answer
verified

Use the Midpoint Rule with four squares of equal size to estimate the double integral. $\iint _ { R } \cos \left( x ^ { 4 } + y ^ { 4 } \right) d A , R = \{ ( x , y ) \mid 0 \leq x \leq 0.5,0 \leq y \leq 0.5 \}$

Correct Answer
verified

Calculate the iterated integral. $\int _ { 1 } ^ { 4 } \int _ { 0 } ^ { 3 } ( 3 + 4 x y ) d x d y$

Correct Answer
verified

Calculate the iterated integral. $\int _ { 0 } ^ { 1 } \int _ { 0 } ^ { y } 5 \cos \left( y ^ { 2 } \right) d x d y$

Correct Answer
verified

Use spherical coordinate to find the volume above the cone $z ^ { 2 } = x ^ { 2 } + y ^ { 2 }$ and inside sphere $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 2 a z$ .

Correct Answer
verified

Exam 15: Multiple Integrals

Use spherical coordinates. Evaluate ∭B(x2+y2+z2) 2dV\iiint _ { B } \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { 2 } d V∭B​(x2+y2+z2) 2dV , where BBB is the ball with center the origin and radius 4 .

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Find the center of mass of the lamina of the region shown if the density of the circular lamina is four times that of the rectangular lamina.

Correct AnswerverifiedShow Answer

Calculate the double integral. ∬R(9x2y3−15x4) dA,R={(x,y) ∣0≤x≤1,0≤y≤4}\iint _ { R } \left( 9 x ^ { 2 } y ^ { 3 } - 15 x ^ { 4 } \right) d A , R = \{ ( x , y ) \mid 0 \leq x \leq 1,0 \leq y \leq 4 \}∬R​(9x2y3−15x4) dA,R={(x,y) ∣0≤x≤1,0≤y≤4}

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Evaluate the double integral by first identifying it as the volume of a solid. ∬R(15−2x) dA,R={(x,y) ∣2≤x≤5,3≤y≤8}\iint _ { R } ( 15 - 2 x ) d A , R = \{ ( x , y ) \mid 2 \leq x \leq 5,3 \leq y \leq 8 \}∬R​(15−2x) dA,R={(x,y) ∣2≤x≤5,3≤y≤8}

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

Use polar coordinates to find the volume of the solid under the paraboloid z=x2+y2z = x ^ { 2 } + y ^ { 2 }z=x2+y2 and above the disk x2+y2≤9x ^ { 2 } + y ^ { 2 } \leq 9x2+y2≤9 .

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Evaluate the integral by changing to polar coordinates. ∬De−x2−y2dA\iint _ { D } e ^ { - x ^ { 2 } - y ^ { 2 } } d A∬D​e−x2−y2dA DDD is the region bounded by the semicircle x=9−y2x = \sqrt { 9 - y ^ { 2 } }x=9−y2​ and the yyy -axis.

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Calculate the double integral. ∬R(6x2y3−10x4) dA,R={(x,y) ∣0≤x≤1,0≤y≤4}\iint _ { R } \left( 6 x ^ { 2 } y ^ { 3 } - 10 x ^ { 4 } \right) d A , R = \{ ( x , y ) \mid 0 \leq x \leq 1,0 \leq y \leq 4 \}∬R​(6x2y3−10x4) dA,R={(x,y) ∣0≤x≤1,0≤y≤4}

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Calculate the iterated integral.Round your answer to two decimal places. ∫06∫034x+4ydxdy\int _ { 0 } ^ { 6 } \int _ { 0 } ^ { 3 } \sqrt { 4 x + 4 y } d x d y∫06​∫03​4x+4y​dxdy

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Find the Jacobian of the transformation. x=5αsin⁡β,y=4αcos⁡βx = 5 \alpha \sin \beta , y = 4 \alpha \cos \betax=5αsinβ,y=4αcosβ

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Use cylindrical coordinates to evaluate ∭Tx2+y2dV\iiint _ { T } \sqrt { x ^ { 2 } + y ^ { 2 } } d V∭T​x2+y2​dV , where TTT is the solid bounded by the cylinder x2+y2=1x ^ { 2 } + y ^ { 2 } = 1x2+y2=1 and the planes z=4z = 4z=4 and z=8z = 8z=8 .

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Find the area of the surface. The part of the surface z=25−x2−y2z = 25 - x ^ { 2 } - y ^ { 2 }z=25−x2−y2 that lies above the xyx yxy -plane.

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Evaluate the double integral by first identifying it as the volume of a solid. ∬R(15−2x) dA,R={(x,y) ∣3≤x≤7,3≤y≤5}\iint _ { R } ( 15 - 2 x ) d A , R = \{ ( x , y ) \mid 3 \leq x \leq 7,3 \leq y \leq 5 \}∬R​(15−2x) dA,R={(x,y) ∣3≤x≤7,3≤y≤5}

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Sketch the solid bounded by the graphs of the equations z=x2+y2z = x ^ { 2 } + y ^ { 2 }z=x2+y2 and z=50−x2−y2z = 50 - x ^ { 2 } - y ^ { 2 }z=50−x2−y2 , and then use a triple integral to find the volume of the solid.

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

Calculate the iterated integral. ∫14∫03(3+4xy)dxdy\int _ { 1 } ^ { 4 } \int _ { 0 } ^ { 3 } ( 3 + 4 x y ) d x d y∫14​∫03​(3+4xy)dxdy

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Calculate the iterated integral. ∫01∫0y5cos⁡(y2)dxdy\int _ { 0 } ^ { 1 } \int _ { 0 } ^ { y } 5 \cos \left( y ^ { 2 } \right) d x d y∫01​∫0y​5cos(y2)dxdy

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Use spherical coordinate to find the volume above the cone z2=x2+y2z ^ { 2 } = x ^ { 2 } + y ^ { 2 }z2=x2+y2 and inside sphere x2+y2+z2=2azx ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 2 a zx2+y2+z2=2az .

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Use spherical coordinates. Evaluate $\iiint _ { B } \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { 2 } d V$ , where $B$ is the ball with center the origin and radius 4 .

Correct Answer
verified

Correct Answer
verified

Calculate the double integral. $\iint _ { R } \left( 9 x ^ { 2 } y ^ { 3 } - 15 x ^ { 4 } \right) d A , R = \{ ( x , y ) \mid 0 \leq x \leq 1,0 \leq y \leq 4 \}$

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Evaluate the double integral by first identifying it as the volume of a solid. $\iint _ { R } ( 15 - 2 x ) d A , R = \{ ( x , y ) \mid 2 \leq x \leq 5,3 \leq y \leq 8 \}$

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Use polar coordinates to find the volume of the solid under the paraboloid $z = x ^ { 2 } + y ^ { 2 }$ and above the disk $x ^ { 2 } + y ^ { 2 } \leq 9$ .

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Evaluate the integral by changing to polar coordinates. $\iint _ { D } e ^ { - x ^ { 2 } - y ^ { 2 } } d A$ $D$ is the region bounded by the semicircle $x = \sqrt { 9 - y ^ { 2 } }$ and the $y$ -axis.

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Calculate the double integral. $\iint _ { R } \left( 6 x ^ { 2 } y ^ { 3 } - 10 x ^ { 4 } \right) d A , R = \{ ( x , y ) \mid 0 \leq x \leq 1,0 \leq y \leq 4 \}$

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Calculate the iterated integral.Round your answer to two decimal places. $\int _ { 0 } ^ { 6 } \int _ { 0 } ^ { 3 } \sqrt { 4 x + 4 y } d x d y$

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Find the Jacobian of the transformation. $x = 5 \alpha \sin \beta , y = 4 \alpha \cos \beta$

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Use cylindrical coordinates to evaluate $\iiint _ { T } \sqrt { x ^ { 2 } + y ^ { 2 } } d V$ , where $T$ is the solid bounded by the cylinder $x ^ { 2 } + y ^ { 2 } = 1$ and the planes $z = 4$ and $z = 8$ .

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Find the area of the surface. The part of the surface $z = 25 - x ^ { 2 } - y ^ { 2 }$ that lies above the $x y$ -plane.

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Evaluate the double integral by first identifying it as the volume of a solid. $\iint _ { R } ( 15 - 2 x ) d A , R = \{ ( x , y ) \mid 3 \leq x \leq 7,3 \leq y \leq 5 \}$

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Sketch the solid bounded by the graphs of the equations $z = x ^ { 2 } + y ^ { 2 }$ and $z = 50 - x ^ { 2 } - y ^ { 2 }$ , and then use a triple integral to find the volume of the solid.

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Calculate the iterated integral. $\int _ { 1 } ^ { 4 } \int _ { 0 } ^ { 3 } ( 3 + 4 x y ) d x d y$

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Calculate the iterated integral. $\int _ { 0 } ^ { 1 } \int _ { 0 } ^ { y } 5 \cos \left( y ^ { 2 } \right) d x d y$

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Use spherical coordinate to find the volume above the cone $z ^ { 2 } = x ^ { 2 } + y ^ { 2 }$ and inside sphere $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 2 a z$ .

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