The volume of the solid obtained by rotating the region enclosed by y=x^(2),quad x=y^(2) about the line x=-4 can be computed using the method of disks or washers via an integral V=int_(a)^(b) " ? "✓ with limits of integration a= and b= The volume is V= cubic units.

Question

The volume of the solid obtained by rotating the region enclosed by  y=x^(2),quad x=y^(2) about the line  x=-4 can be computed using the method of disks or washers via an integral  V=int_(a)^(b)  " ? "✓ with limits of integration  a= and  b= The volume is  V= cubic units.

Bytelearn · Accepted Answer

Intersection Points: To find the volume of the solid obtained by rotating the region enclosed by the given curves about the line x=-4, we need to set up the integral using the method of disks or washers. First, we need to find the points of intersection of the curves y=x^2 and x=y^2 to determine the limits of integration.Expressing Volume Integral: To find the points of intersection, we set y=x^2 equal to x=y^2, which gives us x^2=y^2. Taking the square root of both sides, we get x=y or x=-y. Since we are dealing with volumes and the region must be in the first quadrant, we only consider x=y. Substituting y for x in y=x^2, we get x=x^4. Solving for x, we find that x=0 or x=1. These are the points of intersection and thus our limits of integration.Simplifying Integrands: Now we need to express the volume integral. Since we are rotating around the line x=-4, we need to adjust our functions to account for the distance from this line. The outer radius R will be the distance from x=-4 to the curve x=y^2, and the inner radius r will be the distance from x=-4 to the curve y=x^2.Integrating Term by Term: The outer radius R is given by R=(-4)-y^2, and the inner radius r is given by r=(-4)-x^2. Since we are using y as our variable of integration, we need to express x in terms of y for the inner radius. From the curve y=x^2, we have x=sqrt(y). So, r=(-4)-sqrt(y).Evaluating Integral: The volume integral using the washer method is given by V=int_a^b π(R^2-r^2) dy, where R is the outer radius and r is the inner radius. Substituting our expressions for R and r, we get V=int_0^1 π(((-4)-y^2)^2-((-4)-sqrt(y))^2) dy.Calculating Numerical Value: Now we simplify the integrand before integrating. We have V=int_0^1 π((16+8y^2+y^4)-(16+8sqrt(y)+y)) dy.Final Volume Calculation: Further simplifying the integrand, we get V=int_0^1 π(8y^2+y^4-8sqrt(y)-y) dy. This simplifies to V=int_0^1 π(y^4+8y^2-8sqrt(y)-y) dy.Now we integrate term by term. The integral of y^4 with respect to y is (1/5)y^5, the integral of 8y^2 is (8/3)y^3, the integral of -8sqrt(y) is -16y^(3/2), and the integral of -y is -(1/2)y^2. So, we have V=π[(1/5)y^5+(8/3)y^3-16y^(3/2)-(1/2)y^2] from 0 to 1.Evaluating the integral from 0 to 1, we get V=π[(1/5)(1)^5+(8/3)(1)^3-16(1)^(3/2)-(1/2)(1)^2] - π[(1/5)(0)^5+(8/3)(0)^3-16(0)^(3/2)-(1/2)(0)^2].Simplifying the expression, we get V=π[(1/5)+(8/3)-16-(1/2)] - π[0]. This simplifies to V=π[(1/5)+(8/3)-16-(1/2)].Calculating the numerical value, we get V=π[(3/15)+(40/15)-(240/15)-(7.5/15)]. This simplifies to V=π[-197.5/15].Finally, we get V=π[-13.1666667]. Since volume cannot be negative, we must have made a mistake in our calculation. Let's go back and check our work.

Full solution

More problems from Find indefinite integrals using the substitution and by parts

Related topics