The base of a solid is the region bounded by y=x^2 and y=4 . Find the volume of th: solid given

The base of a solid is the region bounded by y=x^2 and y=4 ....

Key Concepts

Relating Base Geometry to Cross Sections

This concept links the geometry of the base of the solid to the dimensions of its cross sections. In many problems, the size or dimensions of the cross-sectional shapes depend on the position along the axis, which is derived from the boundary curves of the base. Understanding how these dimensions change allows one to correctly formulate the expression for the area of each slice.

Solid Slicing Method

This concept involves understanding that the volume of a solid with known cross sections can be computed by integrating the areas of those cross sections along an axis. When a solid has slices perpendicular to an axis with a known area function A(x), the total volume is obtained by integrating A(x) over the appropriate bounds. This slicing or disk/washer method is fundamental in solving volume problems in calculus.

Determining Integration Limits

This key idea involves setting the correct limits of integration based on the region over the base of the solid. Such limits are usually determined by finding the intersection points of the curves that bound the region. Correctly identifying these limits ensures the integration covers the entire base of the solid and not more or less.

Geometric Area Formulas

A crucial part of solving these problems is the ability to compute the area of cross sections that are common geometric shapes such as squares, semicircles, and equilateral triangles. Familiarity with these area formulas allows one to express the cross-sectional area function in terms of the variable of integration, which is necessary to set up and evaluate the integral.

Transcript

00:01 In this question, it is given that the base of a solid is the region bounded by the parabola, that is y equals to x square and the line y equals to 4.

00:15 So we need to find out the volume of the solid given that the cross section is perpendicular to the x -axis and the cross -section is in the form in the first part it is in the form squares.

00:29 In second part it is in the form semi circles.

00:33 In third part it is in the form equilateral triangles.

00:37 So starting from the first part, we will find out the volume of that solid region.

00:42 First of all, we should plot a figure for the given base.

00:47 And then we will imagine the solid as given for the different parts.

00:54 So we plotted the graph for the equation that is y equals to x square and y equals to four.

00:59 And we shaded the region enclosed by these two figures now if we consider the cross section in this given figure when we will imagine this as a 3d figure then if we consider the squares which are perpendicular to x -axis then the square will have a side which is that is parallel to this y -axis and that side will be equal to 4 minus that is the length of y here on this in this graph that is y will be like here and this is for 4 minus y will be the side of that square therefore we can find out the volume of that solid as the integral of area dot of d x for the limit x is ranging from minus 2 to 2 here we can see that this region is ranging from x equal to minus 2 to 2 so this will be for the integral minus 2 to 2 and that is area of that square will be a square of that side of that is square so side of square will be uh that is 4 minus x square 4 minus x square as y is here as x square so 4 minus y will give 4 minus x squared dot d x 4 minus x this is edge and square of that edge will be the area of that square so 4 minus x square dot d x and integral for this area dot d x this one is the area of that square dot d x for the limit minus 2 to 2 will give the volume so the volume will be we can expand this whole square and we will get the integral for the limit minus 2 to 2 and whole square of 4 minus x square will give 16 minus that is 8 x square plus x to the power 4 .d .c.

02:59 Now integrating this, we will get this volume as volume v will be equal to.

03:07 We can write this integral as for the limit 0 to 2.

03:12 Multiplying this integral by 2, we will get volume v as twice of, that is, integral of 16 will be 16x, minus 8x square will be integrated to 8x cube over 3, plus x to the power 4 will be integrated to x to the power 5 over 5.

03:30 Now for the limit 0 to 2 as we have already multiplied this integral by 2.

03:36 So now executing this limit we will get this volume as volume v will be equal to 512 over 15 cubic units.

03:54 So we got the volume of the solid if this solid is having squares as the cross sections...

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