Question 1

Find the area of the region that is bounded by the given curve and lies in the specified sector. $r = 9 \sqrt { \sin 2 \theta } , 0 \leq \theta \leq \pi / 2$

Accepted Answer

The answer of Find the area of the region that...

Question 2

Solve the initial-value problem. $\frac { d r } { d t } + 2 t r - r = 0 , r ( 0 ) = 3$

Accepted Answer

The answer of Solve the initial-value problem. \(\frac { d...

Question 3

Set up an integral that represents the length of the curve. Then use your calculator to find the length correct to four decimal places. $x = t - 2 \sin t , \quad y = t - 2 \cos t , \quad 0 \leq t \leq 4 \pi$

Accepted Answer

A)  $28.7298$ 
B)  $26.7298$ 
C)  $25.7298$ 
D)  $24.7298$ 
E)  $27.7298$ 
A)  $28.7298$ 
B)  $26.7298$ 
C)  $25.7298$ 
D)  $24.7298$ 
E)  $27.7298$

Question 4

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. $y = 5 \left( x ^ { 2 } + 4 \right) , y = 5 \left( 12 - x ^ { 2 } \right) ; \text { about } y = - 5$

Accepted Answer

A)  $9525 \pi$ 
B)  $9575 \pi$ 
C)  $9550 \pi$ 
D)  $9600 \pi$ 
E)  None of these 
A)  $9525 \pi$ 
B)  $9575 \pi$ 
C)  $9550 \pi$ 
D)  $9600 \pi$ 
E)  None of these

Find the area of the region that is bounded by the given curve and lies in the specified sector. $r = 9 \sqrt { \sin 2 \theta } , 0 \leq \theta \leq \pi / 2$

Solve the initial-value problem. $\frac { d r } { d t } + 2 t r - r = 0 , r ( 0 ) = 3$

Set up an integral that represents the length of the curve. Then use your calculator to find the length correct to four decimal places. $x = t - 2 \sin t , \quad y = t - 2 \cos t , \quad 0 \leq t \leq 4 \pi$

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. $y = 5 \left( x ^ { 2 } + 4 \right) , y = 5 \left( 12 - x ^ { 2 } \right) ; \text { about } y = - 5$

Functions and Limits

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Integrals

Applications of Integration

Inverse Functions

Techniques of Integration

Further Applications of Integration

Differential Equations

Parametric Equations and Polar Coordinates

Infinite Sequences and Series

Vectors and the Geometry of Space

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Partial Derivatives

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Vector Calculus

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Exam 18: Final Exam

Find the area of the region that is bounded by the given curve and lies in the specified sector. r=9sin⁡2θ,0≤θ≤π/2r = 9 \sqrt { \sin 2 \theta } , 0 \leq \theta \leq \pi / 2r=9sin2θ​,0≤θ≤π/2

Solve the initial-value problem. drdt+2tr−r=0,r(0)=3\frac { d r } { d t } + 2 t r - r = 0 , r ( 0 ) = 3dtdr​+2tr−r=0,r(0)=3

Set up an integral that represents the length of the curve. Then use your calculator to find the length correct to four decimal places. x=t−2sin⁡t,y=t−2cos⁡t,0≤t≤4πx = t - 2 \sin t , \quad y = t - 2 \cos t , \quad 0 \leq t \leq 4 \pix=t−2sint,y=t−2cost,0≤t≤4π

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=5(x2+4),y=5(12−x2); about y=−5y = 5 \left( x ^ { 2 } + 4 \right) , y = 5 \left( 12 - x ^ { 2 } \right) ; \text { about } y = - 5y=5(x2+4),y=5(12−x2); about y=−5

Functions and Limits

Derivatives

Applications of Differentiation

Integrals

Applications of Integration

Inverse Functions

Techniques of Integration

Further Applications of Integration

Differential Equations

Parametric Equations and Polar Coordinates

Infinite Sequences and Series

Vectors and the Geometry of Space

Vector Functions

Partial Derivatives

Multiple Integrals

Vector Calculus

Second-Order Differential Equations

Filters

Find the area of the region that is bounded by the given curve and lies in the specified sector. $r = 9 \sqrt { \sin 2 \theta } , 0 \leq \theta \leq \pi / 2$

Solve the initial-value problem. $\frac { d r } { d t } + 2 t r - r = 0 , r ( 0 ) = 3$

Set up an integral that represents the length of the curve. Then use your calculator to find the length correct to four decimal places. $x = t - 2 \sin t , \quad y = t - 2 \cos t , \quad 0 \leq t \leq 4 \pi$

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. $y = 5 \left( x ^ { 2 } + 4 \right) , y = 5 \left( 12 - x ^ { 2 } \right) ; \text { about } y = - 5$