We use integration to find the volume of a cone of base radius r and altitude h.

The result is that ri = r ( h - yi ) / h, from which we obtain the expressions for the area and volume of the slice.

• Assuming a standard partition we sum the volume contributions Vi and take the limit, obtaining the indicated integral with the indicated result.

We next consider the solid of revolution obtained by rotating the elllipse in the figure below about the x axis.

• We note that this elllipse fits perfectly inside the indicated rectangle. If you do not understand this, you should review the graph of conic sections.

We consider a section of thickness `dxi near x = xi.

• A side view shows that the radius of the section goes from the x axis to the point (xi, y) on the ellipse.
• Since the coordinates satisfy the equation of elllipse, we have xi^2 / a^2 + y^2 / b^2 = 1.
• The distance from the x axis to this point is y, so the radius of the disk is ri = y.
• We thus have xi^2 / a^2 + ri^2 / b^2 = 1.

Using this expression for ri we obtain the indicated expression for the approximate volume Vi of this disk.

• The approximate Riemann some approaches the integral shown in the fifth line.
• The integral is easily evaluated, yielding the total volume V = 4/3 `pi b^2 a.

Video Clip #01

We find the volume of the solid of revolution form when the curve y = 1 + 2 e^x, 0 < x < 3, is revolved around the x axis.

• The radius of the slice formed at or near x = xi is ri = 1 + 2 e^xi.
• The cross-sectional area and the volume of the slice are easily found in terms of this radius.
• The volume of the solid is thus the Riemann sum, which approaches the integral in the last line.
• The integral is easily evaluated.

We find the volume of the solid line above the region shown, with cross-sections parallel to the y axis forming semicircles.

• The diameter of the semicircle at x = xi is therefore the distance di = (1 + e^xi) from the x axis to the curve.
• The radius ri of this semicircle is half the diameter, as indicated.
• The area of a semicircle is half the area of the circle, so we obtain c.s. area Ai as indicated.