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Daerah R dibatasi oleh y=bx​, y=bx untuk x∈[0,2]. Jika volume benda padat yang didapat dengan memutar R terhadap sumbu x adalah π, maka b=....

Pertanyaan

Daerah R dibatasi oleh y equals b square root of x comma space y equals b x untuk x element of left square bracket 0 comma 2 right square bracket. Jika volume benda padat yang didapat dengan memutar R terhadap sumbu x adalah straight pi, maka b=....

  1. 5

  2. 4

  3. 3

  4. 2

  5. 1

A. Rizky

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Pembahasan

i right parenthesis C a r i space p e r p o t o n g a n space k u r v a space d a n space g a r i s  y equals y  b square root of x equals b x  square root of x equals x  x minus square root of x equals 0  square root of x open parentheses square root of x minus 1 close parentheses equals 0    M a k a  square root of x equals 0 rightwards arrow x equals 0    D a n  square root of x minus 1 equals 0 rightwards arrow square root of x equals 1 rightwards arrow x equals 1    i i right parenthesis G a m b a r k a n space d a e r a h space R space  D a e r a h space R space d i t u n j u k k a n space o l e h space d a e r a h space y a n g space d i a r s i r space b e r i k u t space i n i space colon

T e r l i h a t space b a h w a space k u r v a space d a n space g a r i s space b e r p o t o n g a n space p a d a space t i t i k space x equals 0 space d a n space x equals 1. space space    i i right parenthesis H i t u n g space v o l u m e space d a r i space b e n d a space p a d a t space y a n g space d i d a p a t space d e n g a n space m e m u t a r space d a e r a h space R space t e r h a d a p space s u m b u minus x space  D a e r a h space R space t e r b a g i space m e n j a d i space 2 comma space y a i t u space p a d a space i n t e r v a l space 0 less or equal than x less or equal than 1 space d a n space 1 less than x less or equal than 2. space space    V o l u m e space b e n d a space p a d a t space p a d a space i n t e r v a l space 0 less or equal than x less or equal than 1 space a d a l a h  V subscript 1 equals pi integral from 0 to 1 of open parentheses b square root of x close parentheses squared minus open parentheses b x close parentheses squared d x equals pi integral from 0 to 1 of b squared x minus b squared x squared d x equals pi open open parentheses b squared over 2 x squared minus b squared over 3 x cubed close parentheses close square brackets subscript 0 superscript 1  equals pi open square brackets open parentheses b squared over 2 open parentheses 1 close parentheses squared minus b squared over 3 open parentheses 1 close parentheses cubed close parentheses minus open parentheses b squared over 2 open parentheses 0 close parentheses squared minus b squared over 3 open parentheses 0 close parentheses cubed close parentheses close square brackets equals pi open square brackets open parentheses b squared over 2 minus b squared over 3 close parentheses minus open parentheses 0 minus 0 close parentheses close square brackets equals fraction numerator b squared pi over denominator 6 end fraction    S e d a n g k a n space v o l u m e space b e n d a space p a d a t space p a d a space i n t e r v a l space 1 less than x less or equal than 2 space a d a l a h  V subscript 2 equals pi integral from 1 to 2 of open parentheses b x close parentheses squared minus open parentheses b square root of x close parentheses squared d x equals pi integral from 1 to 2 of b squared x squared minus b squared x d x equals pi open open parentheses b squared over 3 x cubed minus b squared over 2 x squared close parentheses close square brackets subscript 1 superscript 2  equals pi open square brackets open parentheses b squared over 3 open parentheses 2 close parentheses cubed minus b squared over 2 open parentheses 2 close parentheses squared close parentheses minus open parentheses b squared over 3 open parentheses 1 close parentheses cubed minus b squared over 2 open parentheses 1 close parentheses squared close parentheses close square brackets equals pi open square brackets open parentheses fraction numerator 8 b squared over denominator 3 end fraction minus 2 b squared close parentheses minus open parentheses b squared over 3 minus b squared over 2 close parentheses close square brackets  equals pi open square brackets fraction numerator 2 b squared over denominator 3 end fraction minus open parentheses negative b squared over 6 close parentheses close square brackets equals pi open parentheses fraction numerator 2 b squared over denominator 3 end fraction plus b squared over 6 close parentheses equals fraction numerator 5 b squared pi over denominator 6 end fraction    S e h i n g g a space v o l u m e space b e n d a space p a d a t space p a d a space i n t e r v a l space 0 less or equal than x less or equal than 2 space a d a l a h  V subscript 1 plus V subscript 2 equals fraction numerator b squared pi over denominator 6 end fraction plus fraction numerator 5 b squared pi over denominator 6 end fraction equals b squared pi    i i i right parenthesis B a n d i n g k a n space v o l u m e space y a n g space d i d a p a t k a n space  D a r i space p e r h i t u n g a n space d i d a p a t k a n space v o l u m e space y a i t u space b squared pi  D a r i space s o a l space d i d a p a t k a n space v o l u m e space y a i t u space pi. space  M a k a  b squared pi equals pi  b squared equals 1  b equals plus-or-minus 1

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Daerah R dibatasi oleh y=bx​,y=bx, untuk x∈[0,2]. Jika volume benda padat yang didapat dengan memutar R terhadap sumbu x adalah π, maka b = ....

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