Finding volume Find the volume of the solid generated by revolving the region bounded by the x -

Finding volume Find the volume of the solid generated by...

Key Concepts

Integration Techniques

Solving integrals that arise in volumes of revolution often requires various techniques of integration, such as substitution or integration by parts. These techniques help simplify and evaluate integrals involving products of functions, enabling the determination of the exact volume of the rotated region.

Solids of Revolution

This concept involves generating a three?dimensional object by rotating a two?dimensional area about an axis. The process transforms an area into a volume, and its calculation typically requires setting up an integral that sums up all the infinitesimally thin slices or shells created by the rotation.

Cylindrical Shell Method

The cylindrical shell method is an integration technique used to calculate the volume of a solid of revolution, particularly useful when the solid is generated by rotating a region about a vertical axis or an axis parallel to the y-axis. In this method, the volume is computed by integrating the lateral surface area of cylindrical shells whose radii, heights, and thickness vary along the axis of rotation.

Washer/Disk Method

This method is used to determine volumes of solids of revolution by slicing the solid perpendicular to the axis of rotation. When the solid has a hole, the washer method is used, which involves subtracting the volume of the inner solid from the outer solid. This method is often more straightforward when the region is bounded by functions that make expressing the radius of the slices easy.

Adjusting for Non-standard Axis of Rotation

When the axis of rotation is not one of the primary coordinate axes, it becomes necessary to adjust the expressions for the radius of the rotated element. This involves computing the distance from a point in the region to the axis of rotation, which can change the form of the integrand from the standard cases and necessitates careful geometric considerations in setting up the integral.

Transcript

00:01 Column 56 we are supposed to find the volume of a region volume of a solid which is generated by revolving the region which is bounded by x -synx so the graph of x -sinex between 0 to pi is like this the area which is bounded by this x -sine x -axis and this between x at 0 and pi so this is the required region and this is rotated first about y -axis and the second of about x equal to pi let's talk about part a and this is rotated about y axis so i think the shell method will be more comfortable here so we take a small rectangle over here it's like this so it's this will be its radius which is basically x and the height will be basically y so the volume will be 2 pi x y and we have to integrate this i'm sorry this will be 2 pi x and we have to integrate this with respect to x within the limits zero to pi we have the value of y as x sine x so let's replace this this will become 2 pi x x x sine x that will be x square sine x t x and we have to integrate this within the limit zero to pi so over here we have to use integration by parts so this is part one this is part two so pi remains outside first term is as it is integration of second term will be minus cos x minus integral of differentiation of first term is 2x integral is minus of cos x and this is further integrated so this becomes 2 pi minus x square cos x plus 2 integral of x x x dx once again integral by integration by parts first term second term and let's open up the brackets as well so this becomes minus if i click let's keep my 2 pi outside and let's open the brackets in the next term.

02:21 So this will become first term as it is integral of second term is sine x minus integral of differentiation of first term is x or differentiation of first term which is x is 1.

02:32 Integration of cause will be sign and this should be once again integrated.

02:38 So this becomes 2 pi minus x square cost x plus 2x, plus 2x, sine x, minus integration of sign will be minus cos so this will be cause and here now we have to put the limits between 0 so 2 pi is outside cost 2 pi is minus 1 so minus pi square minus 1 sign pi is 0 and cost pi is once again minus 1 this is the upper limit minus lower limit will be 0 0 cause 0 is 1 so when we simplify this further we get 2 pi is outside this will be pi square minus 2 minus 2.

03:28 So this will be pi square minus 4.

03:31 So the answer is 2 pi, pi square minus 4.

03:34 So this is the answer of part a.

03:37 Let's talk about part b.

03:39 Now the excess of rotation is changed.

03:41 The excess of rotation now is x is equal to pi.

03:45 So this is the excess of rotation...

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