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Set Up an Integral for the Area of the Surface y=x3,0x2;x-axis y = x ^ { 3 } , 0 \leq x \leq 2 ; x \text {-axis }

Question 69

Multiple Choice

Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis.
- y=x3,0x2;x-axis y = x ^ { 3 } , 0 \leq x \leq 2 ; x \text {-axis }


A) 2π02x31+9x4dx2 \pi \int _ { 0 } ^ { 2 } x ^ { 3 } \sqrt { 1 + 9 x ^ { 4 } } d x
B) π02x21+9x4dx\pi \int _ { 0 } ^ { 2 } x ^ { 2 } \sqrt { 1 + 9 x ^ { 4 } } d x
C) π02x31+9x4dx\pi \int _ { 0 } ^ { 2 } x ^ { 3 } \sqrt { 1 + 9 x ^ { 4 } } d x
D) 2π02x21+9x4dx2 \pi \int _ { 0 } ^ { 2 } x ^ { 2 } \sqrt { 1 + 9 x ^ { 4 } } d x

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