Area in Polar Coordinates & Solid of Revolution Problems

Area in Polar Coordinates Volume of a Solid of Revolution Group 2 Area in Polar Coordinates Find the area of the region inside 𝒓 = 𝟐 and to the right of 𝒓 = 𝒔𝒆𝒄𝜽. 𝟏 𝜷 𝑨= = 𝟐 𝜶 𝟏 𝟐 𝝅 𝟑 𝝅 − 𝟑 (𝒇 𝜽 )𝟐 − (𝒈 𝜽 )𝟐 𝒅𝜽 𝟐 𝟐 − 𝒔𝒆𝒄𝜽 𝟐 𝒅𝜽 𝝅 𝟑 𝟏 = 𝟐( ) 𝟒 − 𝒔𝒆𝒄𝟐 𝜽 𝒅𝜽 𝟐 𝟎 = 𝝅 𝟑 𝟎 =𝟒 = 𝝅 𝝅 − 𝒕𝒂𝒏( ) 𝟑 𝟑 𝟒𝝅 − 𝟑 𝟑 𝟒 − 𝒔𝒆𝒄𝟐 𝜽 𝒅𝜽 𝝅 = [𝟒𝜽 − 𝒕𝒂𝒏𝜽] 𝟑 𝟎 𝟐 = 𝒔𝒆𝒄𝜽 𝟐= 𝟏 𝒄𝒐𝒔(𝜽) 𝟐𝒄𝒐𝒔 𝜽 = 𝟏 𝒄𝒐𝒔 𝜽 = 𝜽= 𝝅 𝝅 𝟓𝝅 𝟑 ( , 𝟎) 𝟑 , 𝟑 𝟏 𝟐 = 𝝅 𝟑 ,− 𝝅 𝟑 = Area in Polar Coordinates Find the area of the region inside 𝒓 = 𝟐 and to the right of 𝒓 = 𝒔𝒆𝒄𝜽. 𝟏 𝜷 𝑨= = 𝟐 𝜶 𝟏 𝟐 𝝅 𝟑 𝝅 − 𝟑 (𝒇 𝜽 )𝟐 − (𝒈 𝜽 )𝟐 𝒅𝜽 𝟐 𝟐 − 𝒔𝒆𝒄𝜽 𝟐 𝒅𝜽 𝝅 𝟑 𝟏 = 𝟐( ) 𝟒 − 𝒔𝒆𝒄𝟐 𝜽 𝒅𝜽 𝟐 𝟎 = 𝝅 𝟑 𝟎 =𝟒 = 𝝅 𝝅 − 𝒕𝒂𝒏( ) 𝟑 𝟑 𝟒𝝅 − 𝟑 𝟑 𝟒 − 𝒔𝒆𝒄𝟐 𝜽 𝒅𝜽 𝝅 = [𝟒𝜽 − 𝒕𝒂𝒏𝜽] 𝟑 𝟎 𝟐 = 𝒔𝒆𝒄𝜽 𝟐= 𝟏 𝒄𝒐𝒔(𝜽) 𝟐𝒄𝒐𝒔 𝜽 = 𝟏 𝒄𝒐𝒔 𝜽 = 𝜽= 𝝅 𝝅 𝟓𝝅 𝟑 ( , 𝟎) 𝟑 , 𝟑 𝟏 𝟐 = 𝝅 𝟑 ,− 𝝅 𝟑 = Find the area of the region enclosed by 𝒓𝟐 = 𝟒𝒄𝒐𝒔(𝟐𝜽). 𝝅 𝟐 𝟏 𝜷 𝟐 𝑨= 𝒓 𝒅𝜽 𝟐 𝜶 𝒅𝒖 =𝟖 𝒄𝒐𝒔(𝒖) 𝟐 𝟎 𝝅 𝟏 𝟒 = 𝟒𝒄𝒐𝒔 𝟐𝜽 𝟐 𝟎 𝝅 𝒅𝜽 𝟏 𝟒 = 𝟒( ) 𝟒𝒄𝒐𝒔(𝟐𝜽) 𝒅𝜽 𝟐 𝟎 =𝟖 𝝅 𝟒 𝟎 𝝅 = 𝟒[𝒔𝒊𝒏(𝒖)] 𝟐 𝟎 𝝅 = 𝟒[𝒔𝒊𝒏( )] 𝟐 = 𝟒(𝟏) 𝒄𝒐𝒔(𝟐𝜽) 𝒅𝜽 =𝟒 𝟎𝟐 = 𝟒𝒄𝒐𝒔(𝟐𝜽) 𝟎 𝟒𝒄𝒐𝒔(𝟐𝜽) = 𝟒 𝟒 𝟎 = 𝒄𝒐𝒔(𝟐𝜽) 𝝅 𝟑𝝅 𝟐𝜽 = 𝟐𝜽 = 𝟐 𝟐 𝝅 𝟑𝝅 𝜽=( , ) 𝟒 𝟒 𝒖 = 𝟐𝜽 𝒅𝒖 = 𝟐𝒅𝜽 𝒅𝒖 𝟐𝒅𝜽 = 𝟐 𝟐 Find the area of the region enclosed by 𝒓𝟐 = 𝟒𝒄𝒐𝒔(𝟐𝜽). 𝝅 𝟐 𝟏 𝜷 𝟐 𝑨= 𝒓 𝒅𝜽 𝟐 𝜶 𝒅𝒖 =𝟖 𝒄𝒐𝒔(𝒖) 𝟐 𝟎 𝝅 𝟏 𝟒 = 𝟒𝒄𝒐𝒔 𝟐𝜽 𝟐 𝟎 𝝅 𝒅𝜽 𝟏 𝟒 = 𝟒( ) 𝟒𝒄𝒐𝒔(𝟐𝜽) 𝒅𝜽 𝟐 𝟎 =𝟖 𝝅 𝟒 𝟎 𝝅 = 𝟒[𝒔𝒊𝒏(𝒖)] 𝟐 𝟎 𝝅 = 𝟒[𝒔𝒊𝒏( )] 𝟐 = 𝟒(𝟏) 𝒄𝒐𝒔(𝟐𝜽) 𝒅𝜽 =𝟒 𝟎𝟐 = 𝟒𝒄𝒐𝒔(𝟐𝜽) 𝟎 𝟒𝒄𝒐𝒔(𝟐𝜽) = 𝟒 𝟒 𝟎 = 𝒄𝒐𝒔(𝟐𝜽) 𝝅 𝟑𝝅 𝟐𝜽 = 𝟐𝜽 = 𝟐 𝟐 𝝅 𝟑𝝅 𝜽=( , ) 𝟒 𝟒 𝒖 = 𝟐𝜽 𝒅𝒖 = 𝟐𝒅𝜽 𝒅𝒖 𝟐𝒅𝜽 = 𝟐 𝟐 Find the area of the region enclosed by 𝒓 = 𝟔𝒔𝒊𝒏(𝜽). 𝜷 𝑨= 𝟏 𝒓𝟐 𝒅𝜽 𝟐 𝜶 = 𝟏𝟖 𝝅 = 𝟑𝟔𝒔𝒊𝒏𝟐 𝜽 𝒅𝜽 = 𝟏𝟖 𝝅 𝟐 𝒄𝒐𝒔 𝟐𝜽 𝒅𝜽 𝟎 𝝅 𝝅 𝟐 𝒅𝒖 = 𝟏𝟖 𝜽 𝟐 − 𝟏𝟖 𝒄𝒐𝒔(𝒖) 𝟐 𝟎 𝟎 𝝅 𝝅 𝟐 𝟐 𝝅 = 𝟏𝟖 − 𝟗 𝒄𝒐𝒔 𝒖 𝒅𝒖 𝟐 𝟎 𝝅 = 𝟗𝝅 − 𝟗[𝒔𝒊𝒏(𝒖)] 𝟎 = 𝟗𝝅 − 𝟗[𝒔𝒊𝒏 𝝅 ] 𝟎 = 𝟗𝝅 − 𝟎 𝟎 = 𝟑𝟔 𝒅𝜽 − 𝟏𝟖 𝟎 𝟏 𝟐 = 𝟐( ) [𝟔𝒔𝒊𝒏 𝜽 𝟐 ] 𝒅𝜽 𝟐 𝟎 𝝅 𝟐 𝝅 𝟐 𝝅 𝟐𝟏 𝟎 𝟐 [𝟏 − 𝒄𝒐𝒔 𝟐𝜽 ] 𝒅𝜽 [𝟏 − 𝒄𝒐𝒔 𝟐𝜽 ] 𝒅𝜽 = 𝟗𝝅 𝒖 = 𝟐𝜽 𝒅𝒖 𝟐𝒅𝜽 = 𝟐 𝟐 Find the area of the region enclosed by 𝒓 = 𝟔𝒔𝒊𝒏(𝜽). 𝜷 𝑨= 𝟏 𝒓𝟐 𝒅𝜽 𝟐 𝜶 = 𝟏𝟖 𝝅 = 𝟑𝟔𝒔𝒊𝒏𝟐 𝜽 𝒅𝜽 = 𝟏𝟖 𝝅 𝟐 𝒄𝒐𝒔 𝟐𝜽 𝒅𝜽 𝟎 𝝅 𝝅 𝟐 𝒅𝒖 = 𝟏𝟖 𝜽 𝟐 − 𝟏𝟖 𝒄𝒐𝒔(𝒖) 𝟐 𝟎 𝟎 𝝅 𝝅 𝟐 𝟐 𝝅 = 𝟏𝟖 − 𝟗 𝒄𝒐𝒔 𝒖 𝒅𝒖 𝟐 𝟎 𝝅 = 𝟗𝝅 − 𝟗[𝒔𝒊𝒏(𝒖)] 𝟎 = 𝟗𝝅 − 𝟗[𝒔𝒊𝒏 𝝅 ] 𝟎 = 𝟗𝝅 − 𝟎 𝟎 = 𝟑𝟔 𝒅𝜽 − 𝟏𝟖 𝟎 𝟏 𝟐 = 𝟐( ) [𝟔𝒔𝒊𝒏(𝜽)]𝟐 𝒅𝜽 𝟐 𝟎 𝝅 𝟐 𝝅 𝟐 𝝅 𝟐𝟏 𝟎 𝟐 [𝟏 − 𝒄𝒐𝒔 𝟐𝜽 ] 𝒅𝜽 [𝟏 − 𝒄𝒐𝒔 𝟐𝜽 ] 𝒅𝜽 = 𝟗𝝅 𝒖 = 𝟐𝜽 𝒅𝒖 𝟐𝒅𝜽 = 𝟐 𝟐 Find the area of the region enclosed by one petal of the rose curve represented by 𝒓 = 𝟐𝒔𝒊𝒏(𝟒𝜽). 𝟏 𝜷 𝟐 𝑨= 𝒓 𝒅𝜽 𝟐 𝜶 = 𝝅 𝟒 𝟏 = [𝟒𝒔𝒊𝒏𝟐 𝟒𝜽 ] 𝒅𝜽 𝟐 𝟎 = 𝟐 𝝅 𝟒 𝟐 𝒔𝒊𝒏 𝟒𝜽 𝒅𝜽 [𝟏 − 𝒄𝒐𝒔(𝟖𝜽)] 𝒅𝜽 𝟎 𝝅 𝟏 𝟒 = [𝟐𝒔𝒊𝒏 𝟒𝜽 ]𝟐 𝒅𝜽 𝟐 𝟎 𝝅 𝟒 = 𝝅 𝟒 𝝅 𝟒 𝒅𝜽 − 𝟎 𝝅 = 𝜽 𝟒 − 𝟎 𝝅 = 𝟒 𝒄𝒐𝒔 𝟖𝜽 𝒅𝜽 𝟎 𝝅 𝟒 𝒄𝒐𝒔 𝒖 𝟎 𝟐𝝅 − 𝒄𝒐𝒔 𝒖 𝒅𝒖 = 𝟎 = 𝟐𝒔𝒊𝒏(𝟒𝜽) 𝝅 𝟎 𝟐𝒔𝒊𝒏(𝟒𝜽) 𝜽 = (𝟎, ) = 𝟒 𝟐 𝟐 𝟒𝜽 = 𝝅 𝟒𝜽 = 𝟎 𝟎 𝝅 𝟐𝝅 − [𝒔𝒊𝒏(𝒖)] 𝝅 𝟒 𝟎 𝟒 𝟏 = 𝟐 [𝟏 − 𝒄𝒐𝒔(𝟖𝜽)] 𝒅𝜽 = 𝝅 − 𝒔𝒊𝒏(𝟐𝝅) 𝟎 𝟐 𝟒 𝟎 𝒅𝒖 𝟖 𝝅 − 𝟎 𝟒 𝝅 = 𝟒 = 𝒖 = 𝟖𝜽 𝒅𝒖 𝟖 𝒅𝜽 = 𝟖 𝟖 𝒅𝒖 = 𝒅𝜽 𝟖 Find the area of the region enclosed by one petal of the rose curve represented by 𝒓 = 𝟐𝒔𝒊𝒏(𝟒𝜽). 𝟏 𝜷 𝟐 𝑨= 𝒓 𝒅𝜽 𝟐 𝜶 = 𝝅 𝟒 𝟏 = [𝟒𝒔𝒊𝒏𝟐 𝟒𝜽 ] 𝒅𝜽 𝟐 𝟎 = 𝟐 𝝅 𝟒 𝒔𝒊𝒏𝟐 𝟒𝜽 𝒅𝜽 [𝟏 − 𝒄𝒐𝒔(𝟖𝜽)] 𝒅𝜽 𝟎 𝝅 𝟏 𝟒 = [𝟐𝒔𝒊𝒏 𝟒𝜽 ]𝟐 𝒅𝜽 𝟐 𝟎 𝝅 𝟒 = 𝝅 𝟒 𝒅𝜽 − 𝟎 𝒄𝒐𝒔 𝟖𝜽 𝒅𝜽 𝟎 𝝅 = 𝜽 𝟒 − 𝟎 𝝅 = 𝟒 𝝅 𝟒 𝝅 𝟒 𝟎 𝒄𝒐𝒔 𝒖 𝒅𝒖 𝟖 𝟏 𝟐𝝅 − 𝒄𝒐𝒔 𝒖 𝒅𝒖 𝟖 𝟎 𝝅 𝟏 𝟐𝝅 − [𝒔𝒊𝒏(𝒖)] 𝝅 𝟒 𝟖 𝟎 𝟒 𝟏 = 𝟐 [𝟏 − 𝒄𝒐𝒔(𝟖𝜽)] 𝒅𝜽 = 𝝅 − 𝟏 𝒔𝒊𝒏(𝟐𝝅) 𝟐 𝟎 𝟒 𝟖 𝟎 = 𝝅 − 𝟎 𝟒 𝝅 = 𝟒 = 𝟎 = 𝟐𝒔𝒊𝒏(𝟒𝜽) 𝝅 𝟎 𝟐𝒔𝒊𝒏(𝟒𝜽) 𝜽 = (𝟎, ) = 𝟒 𝟐 𝟐 𝟒𝜽 = 𝝅 𝟒𝜽 = 𝟎 𝒖 = 𝟖𝜽 𝒅𝒖 𝟖 𝒅𝜽 = 𝟖 𝟖 𝒅𝒖 = 𝒅𝜽 𝟖 Volume of a Solid of Revolution Washer Method Disk Method Shell Method Formula: Formula: Formula: Around x-axis Around x-axis Vertical Axis of Revolution (y) 𝒃 𝒃 𝟐 𝑽= 𝝅 [𝑹(𝒙)] 𝒅𝒙 𝑽= 𝝅 𝒂 𝑹 𝒙 𝟐 𝒅𝒙 [𝑹(𝒚)] 𝒅𝒚 [𝑹(𝒚)]𝟐 − 𝒓 𝒚 𝑽= 𝝅 𝒄 𝑽 = 𝟐𝝅 𝒓(𝒙)𝒉(𝒙) 𝒅𝒙 Horizontal Axis of Revolution (x) 𝒅 𝟐 𝒃 𝒂 Around y-axis 𝒅 𝒄 − 𝒓 𝒙 𝒂 Around y-axis 𝑽= 𝝅 𝟐 𝟐 𝒅𝒚 𝒅 𝑽 = 𝟐𝝅 𝒓(𝒚)𝒉(𝒚) 𝒅𝒚 𝒄 Thanks !

Area in Polar Coordinates & Solid of Revolution Problems

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