Question

Sketch the region bounded by the curves x=y3,x=1, and y=0 then find the volume of the solid generated by revolving this region about the y-axis.

Answer 1

According to the question, the region is bounded by the curve with the given data as $x = y^{3} , x=1, y=0.$

To compute the volume of the solid, use the disk method. The given object is rotated along the horizontal axis with the disk cross sectional area and the volume can be written as:

$v=\int\limits^a_b {\pi x^{2} } \, dx$

As per question, when x = 1 then the value of y is: y = 1

Now, substituting calculated values in the expression:

$v=\int\limits^a_b {\pi x^{2} } \, dx=v=\int\limits^0_1 {\pi x^{2} } \, dx$

Substitute the value of the variable 'x':

Therefore, the expression can be written as:

$v=\int\limits^a_b {\pi x^{2} } \, dx=v=\int\limits^0_1 {\pi x^{2} } \, dx=\int\limits^0_1 {\pi (y^{3}) ^{2} } \, dx=\frac{\pi }{7}=0.448$

To learn more about the integration from the given link:

brainly.com/question/988162

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