Pertanyaan

Volume benda putar yang terbentuk karena dibatasi oleh kurva y = 2 cos x diputar sejauh 36 0 o mengelilingi sumbu X pada selang 0 ≤ x ≤ 2 π adalah ...

Volume benda putar yang terbentuk karena dibatasi oleh kurva y = 2 cos x diputar sejauh $36 0^{o}$ mengelilingi sumbu X pada selang $0 \leq x \leq \frac{π}{2}$ adalah ...

satuan volume
satuan volume
satuan volume
satuan volume
satuan volume

I. Roy

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Pembahasan

ingat kembali rumus sudut rangkap untuk cos x:

$cos space 2 x equals 2 space cos space squared x minus 1 space space space space space space space space a t a u cos squared x space equals fraction numerator 1 plus cos space 2 x over denominator 2 end fraction m a k a comma v o l u m e equals space straight pi integral subscript 0 superscript straight pi over 2 end superscript open parentheses 2 space cos space straight x close parentheses squared dx space space space space space space space space space space space space space space space space equals space 4 straight pi integral subscript 0 superscript straight pi over 2 end superscript cos squared straight x space dx space space space space space space space space space space space space space space space space space equals space 4 straight pi integral subscript 0 superscript straight pi over 2 end superscript open parentheses fraction numerator 1 plus cos space 2 straight x over denominator 2 end fraction close parentheses dx space space space space space space space space space space space space space space space space space equals 2 straight pi open parentheses straight x plus 1 half sin space 2 straight x left enclose blank with 0 below and straight pi over 2 on top end enclose close parentheses space space space space space space space space space space space space space space space space equals space 2 straight pi open square brackets open parentheses straight pi over 2 minus 1 half sin space straight pi close parentheses minus open parentheses 0 plus 1 half sin space 0 close parentheses close square brackets space space space space space space space space space space space space space space space space equals space 2 straight pi open parentheses straight pi over 2 close parentheses equals straight pi squared satuan space volume$