Daerah D dibatasi oleh garis y = x dan kurva y =x^3. Volume benda putar yang terjadi jika D diputar .... Question from @hoaahm

October 2018 1 27 Report

Daerah D dibatasi oleh garis y = x dan kurva y =x^3. Volume benda putar yang terjadi jika D diputar terhadap sumbu X adalah

izharulhaq78 $\displaystyle \displaystyle \text{tentukan titik potong}\\y_1=y_2\\x=x^3\\0=x^3-x\\0=x^3-x\\0=x(x^2-1)\\0=x(x-1)(x+1)\\x=0\text{ atau }x=1\text{ atau }x=-1\\\\\text{terdapat 2 daerah, yaitu }1\text{ sampai }0\text{ dan }0\text{ sampai }-1\\\\\text{hitung secara terpisah agar lebih aman}$

$\displaystyle V_1=\pi\int\limits^1_0y_1^2-y_2^2\,dx\\V_1=\pi\int\limits^1_0(x)^2-(x^3)^2\,dx\\V_1=\pi\left(\left\lfloor\frac1{2+1}x^{2+1}-\frac1{6+1}x^{6+1}\right\rceil^1_0\right)\\V_1=\pi\left(\left\lfloor\frac1{3}x^{3}-\frac1{7}x^{7}\right\rceil^1_0\right)\\V_1=\pi\left(\frac1{3}(1-0)^{3}-\frac1{7}(1-0)^{7}\right)\\V_1=\pi\left(\frac{7}{21}-\frac{3}{21}\right)\\V_1=\frac4{21}\pi\,\text{sV}$

$\displaystyle V_2=\pi\int\limits^0_{-1}y_1^2-y_2^2\,dx\\V_2=\pi\int\limits^0_{-1}(x)^2-(x^3)^2\,dx\\V_2=\pi\left(\left\lfloor\frac1{2+1}x^{2+1}-\frac1{6+1}x^{6+1}\right\rceil^0_{-1}\right)\\V_2=\pi\left(\left\lfloor\frac1{3}x^{3}-\frac1{7}x^{7}\right\rceil^0_{-1}\right)\\V_2=\pi\left(\frac1{3}(0-(-1))^{3}-\frac1{7}(0-(-1))^{7}\right)\\V_2=\pi\left(\frac1{3}(1)^{3}-\frac1{7}(1)^{7}\right)\\V_2=\pi\left(\frac{7}{21}-\frac3{21}\right)\\V_2=\frac4{21}\pi\,\text{sV}$

$\displaystyle V=V_1+V_2\\V=\frac{4}{21}\pi+\frac{4}{21}\pi\\\boxed{\boxed{V=\frac8{21}\pi\,\text{sV}}}$
$\text{Maaf kalau salah}$

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