Por Exercises 1-4, find the volume under the surface z= (x,9) over the rectangle R. 1. (x.y)= day, R=10,1 10,1) 3. (x, y)-+y", R=(0,1) (0,1) For Exercises 5-12, evalunte the given double integral. 3. LLa-yrdzdy 2. (a.y)=e*,R=10,1)-(-1,1) 4. fta.y)=*+xy+y.R-1,2) - (0,2) LL day+ sinz)dzdy sinzcosty-nidxdy 12. LL 1drdy 8. 10. 13. Let M be a constant. Show that Mdxdy- M(d - eXb -a).
Por Exercises 1-4, find the volume under the surface z= (x,9) over the rectangle R. 1. (x.y)= day, R=10,1 10,1) 3. (x, y)-+y", R=(0,1) (0,1) For Exercises 5-12, evalunte the given double integral. 3. LLa-yrdzdy 2. (a.y)=e*,R=10,1)-(-1,1) 4. fta.y)=*+xy+y.R-1,2) - (0,2) LL day+ sinz)dzdy sinzcosty-nidxdy 12. LL 1drdy 8. 10. 13. Let M be a constant. Show that Mdxdy- M(d - eXb -a).
Essentials Of Investments
11th Edition
ISBN:9781260013924
Author:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.
Publisher:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.
Chapter1: Investments: Background And Issues
Section: Chapter Questions
Problem 1PS
Related questions
Question
![A:IV
>
partial derivative +
V - L "dxdy
Recall that for a general function fx), the integral feldx representa the difference of
the area below the curve y = f(x) but above the r-axis when f(x) 2 0, and the area above the
104
CHAPTER 3. MULTIPLE INTEGRALS
eurve but below the r-axis when f(x)s0. Similarly, the double integral of any continuous
funetion f(x, y) represents the difference of the volume below the surface z=f(x, y) but above
the xy-plane when f(x,y) 20, and the valume above the surface but below the xy-plane when
flz,y) s0. Thus, our method of double integration by means of iterated integrals can be
used to evaluate the double integral of any continuous funetion over a rectangle, regardless
of whether fix, y)20 or not.
Exumple 3.3. Evaluate [ sintx+ y)dxdy.
Solution: Note that f(x,y)= sin(r+y) is both positive and negative over the rectangle [0,)
(0,2n1 We can still evaluate the double integral:
sin(x + yldzdy = (-conlx + y)dy
(-cos(y+)+ cos y)dy
- - sin(y+n)+ sin y. --sin3r + sin 27-(-sinn+ sin0)
Exercises
A
For Exercises 1-4, find the volume under the surface z = ((r,y) over the rectangle R.
2. fls, y)=e*, R=[0,1] -[-1,1]
1. fx, y)=4xy, R=[0,1] [0,1]
3. f(x, y) =x+y*.R=[0,1]x [0,1]
4. fiz,y)=x*+xy + yª, R = [1,2]= [0,2)
For Exercises 5-12, evaluate the given double integral.
z(x+y}dxdy
6.
a+2)dzdy
C[ zy+ sinx)dzdy
7.
9.
xycostxy)dxdy
sinxcosty-)dxdy
11
Idxdy
13. Let M be a constant. Show that Mdxdy = M(d - eXb -a).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffced689a-b531-4020-b130-2a1a196809ca%2F005d031e-d48b-4b45-9c5b-5307db8fa9c6%2Fqpav2wb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A:IV
>
partial derivative +
V - L "dxdy
Recall that for a general function fx), the integral feldx representa the difference of
the area below the curve y = f(x) but above the r-axis when f(x) 2 0, and the area above the
104
CHAPTER 3. MULTIPLE INTEGRALS
eurve but below the r-axis when f(x)s0. Similarly, the double integral of any continuous
funetion f(x, y) represents the difference of the volume below the surface z=f(x, y) but above
the xy-plane when f(x,y) 20, and the valume above the surface but below the xy-plane when
flz,y) s0. Thus, our method of double integration by means of iterated integrals can be
used to evaluate the double integral of any continuous funetion over a rectangle, regardless
of whether fix, y)20 or not.
Exumple 3.3. Evaluate [ sintx+ y)dxdy.
Solution: Note that f(x,y)= sin(r+y) is both positive and negative over the rectangle [0,)
(0,2n1 We can still evaluate the double integral:
sin(x + yldzdy = (-conlx + y)dy
(-cos(y+)+ cos y)dy
- - sin(y+n)+ sin y. --sin3r + sin 27-(-sinn+ sin0)
Exercises
A
For Exercises 1-4, find the volume under the surface z = ((r,y) over the rectangle R.
2. fls, y)=e*, R=[0,1] -[-1,1]
1. fx, y)=4xy, R=[0,1] [0,1]
3. f(x, y) =x+y*.R=[0,1]x [0,1]
4. fiz,y)=x*+xy + yª, R = [1,2]= [0,2)
For Exercises 5-12, evaluate the given double integral.
z(x+y}dxdy
6.
a+2)dzdy
C[ zy+ sinx)dzdy
7.
9.
xycostxy)dxdy
sinxcosty-)dxdy
11
Idxdy
13. Let M be a constant. Show that Mdxdy = M(d - eXb -a).
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