What integral could you use to calculate the volume of the solid generated by rotating the region bounded by x= √[sin(y)], y=0, y=π, x=0 about the line x=4?
I recognise that here y is the independent variable, and the rotation is similar to rotating around the y axis except we are using x=4 instead. So I think the integral will have the form
\(\displaystyle \int_{0}^{\Pi} \Pi f(y)^2 dy
\)
since we are rotating around x=4, the integral is
\(\displaystyle \int_{0}^{\Pi} \Pi (4-\sqrt{siny})^2 dy
\)
This is a multiple choice question and none of the options are as above. The options are:
\(\displaystyle \Pi \int_{0}^{\Pi}16 - (4-\sqrt{siny})^2 dy\)
\(\displaystyle \Pi \int_{0}^{\Pi} (4-y)(\sqrt{siny}) dy\)
\(\displaystyle 2\Pi \int_{0}^{\Pi} (4-y)(siny) dy\)
\(\displaystyle 2\Pi \int_{0}^{\Pi} (y-4)(\sqrt{siny}) dy
\)
I recognise that here y is the independent variable, and the rotation is similar to rotating around the y axis except we are using x=4 instead. So I think the integral will have the form
\(\displaystyle \int_{0}^{\Pi} \Pi f(y)^2 dy
\)
since we are rotating around x=4, the integral is
\(\displaystyle \int_{0}^{\Pi} \Pi (4-\sqrt{siny})^2 dy
\)
This is a multiple choice question and none of the options are as above. The options are:
\(\displaystyle \Pi \int_{0}^{\Pi}16 - (4-\sqrt{siny})^2 dy\)
\(\displaystyle \Pi \int_{0}^{\Pi} (4-y)(\sqrt{siny}) dy\)
\(\displaystyle 2\Pi \int_{0}^{\Pi} (4-y)(siny) dy\)
\(\displaystyle 2\Pi \int_{0}^{\Pi} (y-4)(\sqrt{siny}) dy
\)
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